[proofplan] The only point is to compare the two adjoint operations. Since the entries of $Q$ are real, entrywise complex conjugation fixes $Q$, so the conjugate transpose $Q^*$ of the complex matrix is the same matrix as the real transpose $Q^\top$. The defining equations for orthogonality and unitarity therefore become the same equation. [/proofplan] [step:Identify the conjugate transpose with the real transpose] Write $Q = (q_{ij})_{1 \leq i,j \leq n}$, where $q_{ij} \in \mathbb{R}$ for every $i,j \in \{1,\dots,n\}$. Let $Q^*$ denote the conjugate transpose of $Q$ after regarding $Q$ as a complex matrix. For every $i,j \in \{1,\dots,n\}$, the reality of $q_{ji}$ gives $\overline{q_{ji}} = q_{ji}$, and hence \begin{align*} (Q^*)_{ij} = \overline{q_{ji}} = q_{ji} = (Q^\top)_{ij}. \end{align*} Since the two matrices have the same entries, $Q^* = Q^\top$. [guided] Write $Q = (q_{ij})_{1 \leq i,j \leq n}$, with each entry $q_{ij}$ lying in $\mathbb{R}$. When we regard $Q$ as a complex matrix, we are using the canonical inclusion $\mathbb{R} \subset \mathbb{C}$, so each real entry $q_{ij}$ is viewed as the complex number $q_{ij} + 0i$. The conjugate transpose $Q^*$ is formed by first transposing and then complex-conjugating entries. Thus, for every pair of indices $i,j \in \{1,\dots,n\}$, \begin{align*} (Q^*)_{ij} = \overline{q_{ji}}. \end{align*} Because $q_{ji}$ is real, complex conjugation fixes it: \begin{align*} \overline{q_{ji}} = q_{ji}. \end{align*} Therefore \begin{align*} (Q^*)_{ij} = q_{ji} = (Q^\top)_{ij}. \end{align*} This equality holds for every entry, so the matrices are equal: \begin{align*} Q^* = Q^\top. \end{align*} [/guided] [/step] [step:Translate orthogonality into unitarity] Let $I_n$ denote the $n \times n$ identity matrix, regarded over $\mathbb{R}$ or over $\mathbb{C}$ according to context. By definition, the real matrix $Q$ is orthogonal iff \begin{align*} Q^\top Q = I_n. \end{align*} By definition, the complex matrix $Q$ is unitary iff \begin{align*} Q^*Q = I_n. \end{align*} Using $Q^* = Q^\top$, these two equations are identical. Hence $Q$ is orthogonal iff $Q$ is unitary. [/step]