Unitary conjugation is a channel

The single-operator family $(U)$ is a Kraus representation of ${\mathrm{Ad}}_U$ by [F1]. Hence the map is completely positive by [F2]; its adjoint-square sum is $U^{†}U=I_H$ by [F1], so it is trace-preserving by [F3]. It is a channel by [F4].

The composite $VU$ is surjective and norm-preserving because both $U$ and $V$ are; hence $VU$ is a unitary map by [F1]. For every $X\in \mathcal{L}(H)$, [F1] and [F5] give ${\mathrm{Ad}}_V({\mathrm{Ad}}_U(X))=V(UX{U}^{†})V^{†}=(VU)X(U^{†}{V}^{†})=(VU)X(VU{)}^{†}$. Thus ${\mathrm{Ad}}_{VU}={\mathrm{Ad}}_V\circ {\mathrm{Ad}}_U$. Also $U^{†}$ is a unitary map: for $y=Ux$, [F1] gives $U^{†}y=x$, so $U^{†}$ preserves norms, and it is onto because $U^{†}(Ux)=x$. Applying the composition identity to $U$ and $U^{†}$ in both orders shows that ${\mathrm{Ad}}_{U^{†}}$ is a two-sided inverse. It is a channel by step 1.1 applied to $U^{†}$.