# Interesting properties of maps whose natural representation is unitary?

Let $$\rho \in L(\mathcal{X})$$ be a state in the space of linear operators acting on some complex Hilbert space $$\mathcal{X}$$. I'm interested in linear maps $$\Phi: L(\mathcal{X}) \rightarrow L(\mathcal{X})$$ whose natural representation is unitary, i.e. letting $$U(\mathcal{Y})$$ denote the set of unitaries acting on $$\mathcal{Y}$$ there is some $$K(\Phi) \in U(\mathcal{X} \otimes \mathcal{X})$$ such that

$$\text{Vec}\left({\Phi(\rho)} \right) = K(\Phi) \text{Vec}(\rho)$$

What conditions are necessary/sufficient for $$\Phi$$ to be a channel (completely postive, trace-preserving)?

I would also appreciate references discussing this kind of map, if such a thing exists.

Theorem. Let $$\Phi\in L(\mathbb C^{n\times n})$$ be given such that $$K(\Phi)$$ is unitary. The following statements are equivalent.
1. $$\Phi$$ is a channel
2. $$\Phi$$ is completely positive
3. $$\Phi=U(\cdot)U^*$$ for some unitary $$U$$
Proof. As 3. $$\Rightarrow$$ 1. $$\Rightarrow$$ 2. is trivial we only have to prove "2. $$\Rightarrow$$ 3.". The key here are the well known identities $$K(\Psi\Xi)=K(\Psi)K(\Xi)$$ and $$K(\Psi^\dagger)=K(\Psi)^\dagger$$ which hold for all linear maps $$\Psi,\Xi$$, where $$\Psi^\dagger$$ is the dual map of $$\Psi$$ with respect to the Hilbert-Schmidt inner product $$\langle A,B\rangle_{\rm HS}:={\rm tr}(A^\dagger B)$$ (i.e. $$\Psi^\dagger$$ is the unique linear map which satisfies $$\langle A,\Psi(B)\rangle_{\rm HS}=\langle \Psi^\dagger(A),B\rangle_{\rm HS}$$ for all $$A,B$$). Thus $$K(\Phi)$$ being unitary implies $$K({\rm id})={\bf1}=K(\Phi)^\dagger K(\Phi)=K(\Phi^\dagger\circ\Phi)\,;$$ in particular this shows that $$\Phi$$ is invertible and its inverse $$\Phi^\dagger$$ is completely positive (because $$\Phi$$ is). But this is only possible if the Kraus rank of $$\Phi$$ is $$1$$, which is a special case of Theorem 3.1 in Chapter 2 of Davies' book "Quantum Theory of Open Systems" (1976). This, in turn, means that $$K(\Phi)=K(X(\cdot)X^*)=(X^*)^T\otimes X=\overline{X}\otimes X$$ (in the second step we used a standard vectorization identity) for some $$X\in\mathbb C^{n\times n}$$. Obviously, $$\overline{X}\otimes X$$ is unitary if and only if $$X$$ is unitary which concludes the proof. $$\square$$