[proofplan] The strategy is **averaging**: start from any Hermitian inner product on $V$ and replace it by its average over the action of $G$. Sesquilinearity and positive-definiteness pass through the averaging because they are preserved by each summand (the action of each $g \in G$ is a $\mathbb{C}$-linear bijection). $G$-invariance is then forced by a re-indexing argument: translating the averaging variable by a fixed $h \in G$ permutes $G$ and so leaves the sum unchanged. [/proofplan]

[step:Fix an arbitrary Hermitian inner product on $V$ as a starting point] Pick a basis $\mathbf{e}_1, \dots, \mathbf{e}_n$ of $V$ and define the sesquilinear form \begin{align*} (\cdot, \cdot): V \times V &\to \mathbb{C} \\ \Big( \sum_i a_i \mathbf{e}_i,\; \sum_j b_j \mathbf{e}_j \Big) &\mapsto \sum_{i=1}^n a_i \overline{b_i}. \end{align*} This is a Hermitian inner product on $V$: it is linear in the first argument, conjugate-linear in the second, satisfies $(\mathbf{w}, \mathbf{v}) = \overline{(\mathbf{v}, \mathbf{w})}$, and $(\mathbf{v}, \mathbf{v}) = \sum_i |a_i|^2 \geq 0$ with equality iff all $a_i = 0$, i.e., iff $\mathbf{v} = 0$. There is in general no reason for $(\cdot, \cdot)$ to be $G$-invariant. [/step]

[step:Define the averaged form $\langle \cdot, \cdot \rangle$ and verify sesquilinearity] Define \begin{align*} \langle \cdot, \cdot \rangle: V \times V &\to \mathbb{C} \\ (\mathbf{v}, \mathbf{w}) &\mapsto \frac{1}{|G|} \sum_{g \in G} (g\mathbf{v},\, g\mathbf{w}). \end{align*} The sum is finite because $G$ is finite; the division by $|G|$ is permissible because $\operatorname{char} \mathbb{C} = 0$. We verify sesquilinearity. Fix $\mathbf{w} \in V$ and let $\mathbf{v}_1, \mathbf{v}_2 \in V$, $\alpha, \beta \in \mathbb{C}$. Each $g \in G$ acts $\mathbb{C}$-linearly on $V$ (this is part of the definition of a representation), so $g(\alpha \mathbf{v}_1 + \beta \mathbf{v}_2) = \alpha (g\mathbf{v}_1) + \beta (g\mathbf{v}_2)$. Linearity of $(\cdot, \cdot)$ in its first argument then gives \begin{align*} (g(\alpha \mathbf{v}_1 + \beta \mathbf{v}_2),\, g\mathbf{w}) = \alpha (g\mathbf{v}_1, g\mathbf{w}) + \beta (g\mathbf{v}_2, g\mathbf{w}). \end{align*} Summing over $g \in G$ and dividing by $|G|$, \begin{align*} \langle \alpha \mathbf{v}_1 + \beta \mathbf{v}_2, \mathbf{w} \rangle = \alpha \langle \mathbf{v}_1, \mathbf{w} \rangle + \beta \langle \mathbf{v}_2, \mathbf{w} \rangle. \end{align*} Conjugate-linearity in the second argument is identical, using conjugate-linearity of $(\cdot, \cdot)$ in its second argument. Hermitian symmetry $\langle \mathbf{w}, \mathbf{v} \rangle = \overline{\langle \mathbf{v}, \mathbf{w} \rangle}$ follows by taking conjugates inside the sum. [/step]

[step:Verify positive-definiteness of $\langle \cdot, \cdot \rangle$]Let $\mathbf{v} \in V$. Then \begin{align*} \langle \mathbf{v}, \mathbf{v} \rangle = \frac{1}{|G|} \sum_{g \in G} (g\mathbf{v}, g\mathbf{v}) \geq 0, \end{align*} because each summand $(g\mathbf{v}, g\mathbf{v})$ is non-negative by positive-definiteness of $(\cdot, \cdot)$. Suppose $\langle \mathbf{v}, \mathbf{v} \rangle = 0$. Since each $(g\mathbf{v}, g\mathbf{v}) \geq 0$ and the sum is $0$, every summand vanishes; in particular, taking $g = 1_G$, $(\mathbf{v}, \mathbf{v}) = 0$, so $\mathbf{v} = 0$ by positive-definiteness of $(\cdot, \cdot)$. Hence $\langle \cdot, \cdot \rangle$ is positive-definite.[/step]

[guided]We need both directions of positive-definiteness: $\langle \mathbf{v}, \mathbf{v} \rangle \geq 0$ and $\langle \mathbf{v}, \mathbf{v} \rangle = 0 \implies \mathbf{v} = 0$. For the first, each summand $(g\mathbf{v}, g\mathbf{v}) \geq 0$ because $(\cdot, \cdot)$ is positive-definite. The sum of non-negative reals is non-negative, and dividing by the positive number $|G|$ preserves the sign. For the second, we use the fact that a finite sum of non-negative reals is zero iff each summand is zero. So $\langle \mathbf{v}, \mathbf{v} \rangle = 0$ forces $(g\mathbf{v}, g\mathbf{v}) = 0$ for **every** $g \in G$. We do not need every summand — just one — and the cleanest choice is $g = 1_G$ (the identity element of $G$), for which $g\mathbf{v} = \mathbf{v}$ since $\rho(1_G) = \mathrm{id}_V$. Thus $(\mathbf{v}, \mathbf{v}) = 0$, and positive-definiteness of $(\cdot, \cdot)$ forces $\mathbf{v} = 0$. This is the only place where the inclusion of the identity element in the averaging matters; without it, vectors in the kernel of $\rho$ (if $\rho$ were not faithful) might still produce $\langle \mathbf{v}, \mathbf{v} \rangle = 0$ for non-zero $\mathbf{v}$. As stated, the argument is fine because $1_G \in G$ always.[/guided]

[step:Verify $G$-invariance via re-indexing the sum]Let $h \in G$ and $\mathbf{v}, \mathbf{w} \in V$. Compute \begin{align*} \langle h\mathbf{v}, h\mathbf{w} \rangle = \frac{1}{|G|} \sum_{g \in G} (g(h\mathbf{v}),\, g(h\mathbf{w})) = \frac{1}{|G|} \sum_{g \in G} ((gh)\mathbf{v},\, (gh)\mathbf{w}), \end{align*} where the last equality uses the homomorphism property $\rho(g)\rho(h) = \rho(gh)$. Make the substitution $g' := gh$. The map \begin{align*} G &\to G \\ g &\mapsto gh \end{align*} is a bijection (right multiplication by $h$ has two-sided inverse right multiplication by $h^{-1}$), so as $g$ ranges over $G$, so does $g'$. Therefore \begin{align*} \frac{1}{|G|} \sum_{g \in G} ((gh)\mathbf{v},\, (gh)\mathbf{w}) = \frac{1}{|G|} \sum_{g' \in G} (g'\mathbf{v},\, g'\mathbf{w}) = \langle \mathbf{v}, \mathbf{w} \rangle. \end{align*} Thus $\langle h\mathbf{v}, h\mathbf{w} \rangle = \langle \mathbf{v}, \mathbf{w} \rangle$ for all $h \in G$ and $\mathbf{v}, \mathbf{w} \in V$, which is exactly $G$-invariance.[/step]

[guided]This is the crux of the proof. We must check that for every $h \in G$, \begin{align*} \langle h\mathbf{v}, h\mathbf{w} \rangle = \langle \mathbf{v}, \mathbf{w} \rangle. \end{align*} Unfolding the definition of $\langle \cdot, \cdot \rangle$ on the left, \begin{align*} \langle h\mathbf{v}, h\mathbf{w} \rangle = \frac{1}{|G|} \sum_{g \in G} (g(h\mathbf{v}),\, g(h\mathbf{w})). \end{align*} The action is by group homomorphism, $\rho(g)\rho(h) = \rho(gh)$, so $g(h\mathbf{v}) = (gh)\mathbf{v}$. The sum becomes \begin{align*} \frac{1}{|G|} \sum_{g \in G} ((gh)\mathbf{v},\, (gh)\mathbf{w}). \end{align*} The sum is now over the products $\{gh : g \in G\}$. The map $g \mapsto gh$ from $G$ to itself is a bijection — its inverse is $g \mapsto gh^{-1}$ — so as $g$ ranges over $G$, the product $gh$ also ranges over all of $G$, hitting each element exactly once. Renaming $g' = gh$, \begin{align*} \frac{1}{|G|} \sum_{g \in G} ((gh)\mathbf{v},\, (gh)\mathbf{w}) = \frac{1}{|G|} \sum_{g' \in G} (g'\mathbf{v},\, g'\mathbf{w}) = \langle \mathbf{v}, \mathbf{w} \rangle. \end{align*} So the sum is unchanged. This is the standard "translation invariance of summation over a group": any function $f: G \to \mathbb{C}$ satisfies $\sum_{g \in G} f(gh) = \sum_{g \in G} f(g)$ because right multiplication by $h$ permutes $G$. Averaging is the only natural way to manufacture an invariant out of a non-invariant object using only the group-theoretic data.[/guided]

[step:Conclude] The form $\langle \cdot, \cdot \rangle: V \times V \to \mathbb{C}$ is sesquilinear and positive-definite (so it is a Hermitian inner product) and satisfies $\langle h\mathbf{v}, h\mathbf{w} \rangle = \langle \mathbf{v}, \mathbf{w} \rangle$ for every $h \in G$ and $\mathbf{v}, \mathbf{w} \in V$ (so it is $G$-invariant). This produces the required $G$-invariant Hermitian inner product on $V$. [/step]