This is again a request for references. I'd appreciate a pointer to any published proof of the following:

Proposition. Given $n \in \mathbb{N}^+$, let $\Phi$ be a function $\mathbb{C}^n > \to \mathbb{C}^n$. Then $\Phi$ is an isometry of $(\mathbb{C}^n,\|\cdot\|)$ into itself if and only if there exist a unitary $U \in > \mathbb{C}^{n,n}$ and an orthogonal $O \in \mathbb{R}^{n,n}$ such that $\Phi(z) - \Phi(0) = U\; (\Re(z) + i\; O \; > \Im(z))$ for all $z \in > \mathbb{C}^n$.

Here $\|\cdot\|$ stands for the usual norm $\mathbb{C}^n \times \mathbb{C}^n \to \mathbb{R}: (z_1, \ldots, z_n) \mapsto \left(\sum_{k=1}^n |z_k|^2\right)$, and an isometry of $(\mathbb{C}^n,\|\cdot\|)$ into itself is a function $f: \mathbb{C}^n \to \mathbb{C}^n$ such that $\|f(z)-f(w)\| = \|z-w\|$ for all $z,w \in \mathbb{C}^n$.

This is again a request for references. I'd appreciate a pointer to any published proof of the following:

Proposition. Given $n \in \mathbb{N}^+$, let $\Phi$ be a function $\mathbb{C}^n > \to \mathbb{C}^n$. Then $\Phi$ is an isometry of $(\mathbb{C}^n,\|\cdot\|)$ into itself if and only if there exist a unitary $U \in > \mathbb{C}^{n,n}$ and an orthogonal $O \in \mathbb{R}^{n,n}$ such that $\Phi(z) - \Phi(0) = U\; (\Re(z) + i\; O \; > \Im(z))$ for all $z \in > \mathbb{C}^n$.

Here $\|\cdot\|$ stands for the usual norm $\mathbb{C}^n \times \mathbb{C}^n \to \mathbb{R}: (z_1, \ldots, z_n) \mapsto \left(\sum_{k=1}^n |z_k|^2\right)$, and an isometry of $(\mathbb{C}^n,\|\cdot\|)$ into itself is a function $f: \mathbb{C}^n \to \mathbb{C}^n$ such that $\|f(z)-f(w)\| = \|z-w\|$ for all $z,w \in \mathbb{C}^n$.

Edit. I almost forgot! Thanks in advance for any feedback.