[proofplan] For (1), the **conjugate representation** $\bar{R}: g \mapsto \overline{R(g)}$ (entrywise complex conjugation of matrices) is a homomorphism because conjugation is a ring automorphism of $\mathbb{C}$, and its character is $\overline{\chi}$. Irreducibility transfers because complex conjugation sends $\mathbb{C}G$-submodules to $\mathbb{C}G$-submodules and is a bijection on subspaces. For (2), the twisted map $g \mapsto \varepsilon(g) R(g)$ is a homomorphism because $\varepsilon$ is a homomorphism and scalars commute with matrices, and its character is $\varepsilon \chi$. Irreducibility transfers because multiplying by a nonzero scalar at each $g$ preserves the lattice of $\mathbb{C}G$-invariant subspaces. [/proofplan]

[step:Fix a matrix realization of $\chi$ and define the two candidate representations]Choose a $\mathbb{C}$-basis of the underlying representation space $V$ to identify $V \cong \mathbb{C}^n$ with $n = \dim V$. Write \begin{align*} R: G &\to \operatorname{GL}_n(\mathbb{C}) \\ g &\mapsto R(g) \end{align*} for the matrix representation affording $\chi$, so $\chi(g) = \operatorname{tr}(R(g))$ for every $g \in G$. Define two candidate maps: \begin{align*} \bar{R}: G &\to \operatorname{GL}_n(\mathbb{C}) & R': G &\to \operatorname{GL}_n(\mathbb{C}) \\ g &\mapsto \overline{R(g)} & g &\mapsto \varepsilon(g)\, R(g), \end{align*} where $\overline{R(g)}$ denotes entrywise complex conjugation of the matrix $R(g)$, and $\varepsilon: G \to \mathbb{C}^\times$ is the given linear character (a one-dimensional representation, equivalently a group homomorphism into $\mathbb{C}^\times$). We will check (a) each is a representation, (b) each has the claimed character, and (c) each preserves irreducibility.[/step]

[guided]The strategy is direct: in both parts we exhibit an explicit matrix representation whose trace is the desired function. Then irreducibility transfers because each construction is reversible at the level of subspaces — invariant subspaces match up. For (1), we apply complex conjugation entrywise. Since complex conjugation is a ring automorphism of $\mathbb{C}$ (it preserves $+$, $\cdot$, $0$, $1$, and is a bijection $\mathbb{C} \to \mathbb{C}$), it preserves matrix products and matrix inverses entrywise. For (2), we multiply each matrix $R(g)$ by the scalar $\varepsilon(g)$. Since scalars commute with matrices, $(\varepsilon(g)R(g))(\varepsilon(h)R(h)) = \varepsilon(g)\varepsilon(h) R(g)R(h) = \varepsilon(gh)R(gh)$.[/guided]

[step:Verify $\bar{R}$ is a representation with character $\bar{\chi}$] We check that $\bar{R}$ is a group homomorphism. For $g, h \in G$, since $R$ is a homomorphism, $R(gh) = R(g)R(h)$. Entrywise complex conjugation $\overline{(\,\cdot\,)}$ on matrices satisfies $\overline{AB} = \overline{A} \cdot \overline{B}$ for $A, B \in \operatorname{Mat}_n(\mathbb{C})$ (because $\overline{\sum_k a_{ik}b_{kj}} = \sum_k \overline{a_{ik}}\,\overline{b_{kj}}$ by additivity and multiplicativity of complex conjugation). Hence \begin{align*} \bar{R}(gh) = \overline{R(gh)} = \overline{R(g) R(h)} = \overline{R(g)} \cdot \overline{R(h)} = \bar{R}(g)\, \bar{R}(h). \end{align*} Also $\bar{R}(1) = \overline{R(1)} = \overline{I_n} = I_n$. So $\bar{R}: G \to \operatorname{GL}_n(\mathbb{C})$ is a group homomorphism (the image lies in $\operatorname{GL}_n(\mathbb{C})$ because $\overline{R(g)} \cdot \overline{R(g^{-1})} = \overline{R(g)R(g^{-1})} = \overline{I_n} = I_n$, providing the inverse). The character of $\bar{R}$ at $g$ is \begin{align*} \operatorname{tr}(\bar{R}(g)) = \operatorname{tr}(\overline{R(g)}) = \sum_{i=1}^n \overline{R(g)_{ii}} = \overline{\sum_{i=1}^n R(g)_{ii}} = \overline{\operatorname{tr}(R(g))} = \overline{\chi(g)}, \end{align*} where the third equality uses additivity of complex conjugation on a finite sum. So the character of $\bar{R}$ is $\bar{\chi}$, as claimed. [/step]

[step:Verify $R'$ is a representation with character $\varepsilon \chi$] For $g, h \in G$, \begin{align*} R'(g)\, R'(h) &= (\varepsilon(g)\, R(g))\,(\varepsilon(h)\, R(h)) \\ &= \varepsilon(g)\varepsilon(h)\, R(g)R(h) \quad\text{(scalars commute with matrices)} \\ &= \varepsilon(gh)\, R(gh) \quad\text{(since both } \varepsilon \text{ and } R \text{ are homomorphisms)} \\ &= R'(gh). \end{align*} Also $R'(1) = \varepsilon(1) R(1) = 1 \cdot I_n = I_n$. The image lies in $\operatorname{GL}_n(\mathbb{C})$ because $\varepsilon(g) \in \mathbb{C}^\times$ and $R(g) \in \operatorname{GL}_n(\mathbb{C})$, so their product is invertible with inverse $\varepsilon(g)^{-1} R(g)^{-1} = R'(g^{-1})$. Hence $R': G \to \operatorname{GL}_n(\mathbb{C})$ is a group homomorphism. The character of $R'$ at $g$ is \begin{align*} \operatorname{tr}(R'(g)) = \operatorname{tr}(\varepsilon(g)\, R(g)) = \varepsilon(g)\, \operatorname{tr}(R(g)) = \varepsilon(g)\, \chi(g), \end{align*} using linearity of the trace in the scalar factor. So the character of $R'$ is $\varepsilon \chi$, as claimed. [/step]

[step:Transfer irreducibility from $R$ to $\bar{R}$ when $\chi$ is irreducible]Suppose $\chi$ is irreducible, i.e. $R$ is an irreducible representation: the only $R$-invariant subspaces of $\mathbb{C}^n$ are $\{0\}$ and $\mathbb{C}^n$. We must show that $\bar{R}$ is irreducible. Define entrywise complex conjugation as a map on column vectors: \begin{align*} c: \mathbb{C}^n &\to \mathbb{C}^n \\ (z_1, \ldots, z_n) &\mapsto (\overline{z_1}, \ldots, \overline{z_n}). \end{align*} This is a real-linear (but conjugate-linear over $\mathbb{C}$) bijection. It is **not** $\mathbb{C}$-linear, but the key fact we need is its compatibility with matrices: for any $A \in \operatorname{Mat}_n(\mathbb{C})$ and $v \in \mathbb{C}^n$, \begin{align*} c(A v) = \overline{A}\, c(v). \end{align*} This holds because the $i$-th component is $\overline{\sum_j A_{ij} v_j} = \sum_j \overline{A_{ij}}\, \overline{v_j} = (\overline{A}\, c(v))_i$. Let $W \subseteq \mathbb{C}^n$ be a $\bar{R}$-invariant $\mathbb{C}$-linear subspace. Define $c(W) := \{c(w) : w \in W\}$. Because $c$ is real-linear and conjugate-linear, $c(W)$ is again a $\mathbb{C}$-linear subspace of $\mathbb{C}^n$ (closed under sums; closed under scalar multiplication: $\lambda \cdot c(w) = c(\overline{\lambda} w) \in c(W)$ provided $\overline{\lambda} w \in W$, and since $W$ is a $\mathbb{C}$-subspace with $w \in W$, $\overline{\lambda}w \in W$). Now $c(W)$ is $R$-invariant: for $g \in G$ and $w \in W$, \begin{align*} R(g) \cdot c(w) = c(\overline{R(g)}\, w) = c(\bar{R}(g)\, w) \in c(W), \end{align*} using the compatibility relation in reverse and the $\bar{R}$-invariance of $W$. By irreducibility of $R$, $c(W) \in \{0, \mathbb{C}^n\}$. Since $c$ is a bijection, $W = c^{-1}(c(W)) \in \{0, \mathbb{C}^n\}$. Hence $\bar{R}$ has no nontrivial invariant subspace, i.e. $\bar{R}$ is irreducible. Therefore $\bar{\chi}$ is an irreducible character.[/step]

[guided]We want to show that $\bar{R}$ has no nontrivial invariant subspaces given that $R$ has none. The clean approach is to "transport" subspaces from $\bar{R}$-world to $R$-world via complex conjugation. The bijection we use is entrywise conjugation $c: \mathbb{C}^n \to \mathbb{C}^n$, $c(z) = \bar{z}$. This is a real-linear bijection, but conjugate-linear over $\mathbb{C}$ ($c(\lambda z) = \bar{\lambda}\, c(z)$). Despite not being $\mathbb{C}$-linear, $c$ sends $\mathbb{C}$-subspaces to $\mathbb{C}$-subspaces: if $W \subseteq \mathbb{C}^n$ is closed under addition and complex scalars, then so is $c(W)$ (the conjugate-linearity just reshuffles which scalar multiplication maps to which). The matrix-vector compatibility is $c(Av) = \bar{A}\, c(v)$, which holds entrywise. Equivalently, $A \cdot c(v) = c(\bar{A} v)$. Applying this with $A = R(g)$: \begin{align*} R(g)\, c(w) = c(\overline{R(g)}\, w) = c(\bar{R}(g)\, w). \end{align*} If $w \in W$ and $W$ is $\bar{R}$-invariant, then $\bar{R}(g) w \in W$, so $c(\bar{R}(g)w) \in c(W)$, so $R(g)\, c(w) \in c(W)$. Hence $c(W)$ is $R$-invariant. Irreducibility of $R$ now forces $c(W) = 0$ or $c(W) = \mathbb{C}^n$. Since $c$ is a bijection, we recover $W = 0$ or $W = \mathbb{C}^n$. So $\bar{R}$ is irreducible.[/guided]

[step:Transfer irreducibility from $R$ to $R'$ when $\chi$ is irreducible]Suppose $\chi$ is irreducible. We show $R'(g) = \varepsilon(g) R(g)$ defines an irreducible representation. Crucial observation: for every $g \in G$, the operator $R'(g)$ is a nonzero scalar multiple of $R(g)$. Hence a $\mathbb{C}$-linear subspace $W \subseteq \mathbb{C}^n$ satisfies $R'(g) W \subseteq W$ if and only if $R(g) W \subseteq W$. (Indeed, $R'(g)W = \varepsilon(g) R(g) W$, and multiplication by the nonzero scalar $\varepsilon(g)$ preserves any $\mathbb{C}$-subspace: $\varepsilon(g)W = W$.) Therefore the $R'$-invariant subspaces of $\mathbb{C}^n$ coincide with the $R$-invariant subspaces. Since $R$ is irreducible, the only $R$-invariant subspaces are $\{0\}$ and $\mathbb{C}^n$, so the only $R'$-invariant subspaces are $\{0\}$ and $\mathbb{C}^n$. Hence $R'$ is irreducible, and $\varepsilon \chi$ is an irreducible character.[/step]

[guided]The point is that twisting by a one-dimensional character $\varepsilon$ scales each operator $R(g)$ by a nonzero complex number. Scaling by a nonzero scalar does not change the lattice of invariant subspaces of an operator: if $T$ has invariant subspace $W$ and $c \in \mathbb{C}^\times$, then $(cT)W = c(TW) \subseteq cW = W$ (since $W$ is a $\mathbb{C}$-subspace, multiplication by $c$ maps $W$ into itself, in fact bijectively). Applying this for every $g \in G$: $W$ is $R'(g)$-invariant for all $g$ if and only if $W$ is $R(g)$-invariant for all $g$. So the $R'$-invariant and $R$-invariant subspaces are the same. Irreducibility of $R$ thus implies irreducibility of $R'$. This is why the conclusion is "irreducible if irreducible, reducible if reducible" — the lattice of invariant subspaces is identical for $R$ and $R'$.[/guided]

[step:Collect both constructions] We have shown that $\bar{R}$ and $R'$ are representations of $G$ with characters $\bar{\chi}$ and $\varepsilon \chi$ respectively (Steps 2 and 3), and that each is irreducible whenever $R$ is irreducible (Steps 4 and 5). Hence both $\bar{\chi}$ and $\varepsilon \chi$ are complex characters of $G$, irreducible whenever $\chi$ is. This proves both parts. [/step]