Suppose $\Phi$ is a linear map from $\mathbb{R}^{n\times n}$ to $\mathbb{R}^{n\times n}$, and satisfies that if $A\in \mathbb{R}^{n\times n}$ is positive semidefinite, then $\Phi(A)$ is positive semidefinite. Also $\forall m>1$, let $\mathcal{I}_m$ is an identitiy map on $\mathbb{R}^{m\times m}$, then $\forall A\in\mathbb{R}^{mn\times mn}$, $A$ is positive semidefinite, then $(\Phi\otimes\mathcal{I}_m)A$ is positive semidefinite. $(\Phi\otimes\mathcal{I}_m)A$ here means that $A=\sum_iA_i\otimes B_i$, where $A_i\in\mathbb{R}^{n\times n},B_i\in\mathbb{R}^{m\times m}$, then $(\Phi\otimes\mathcal{I}_m)A=\sum_i\Phi(A_i)\otimes B_i$. The $\otimes$ here means the Kronecker Product. The above conditions is also called completely positive.
My question is can we exchange the position of $\Phi$ and $\mathcal{I}$? That is, $A$ (semidefinite positive) can also be written as $A=\sum_i C_i\otimes D_i$, where $C_i\in\mathbb{R}^{m\times m},D_i\in\mathbb{R}^{n\times n}$, then is $(\mathcal{I}\otimes \Phi)(A)=\sum_i C_i\otimes \Phi(D_i)$ also semidefinite positive?
Attempts: If $A=A_1\otimes B_1$ and $m=n$, it trivially holds because $A_1,B_1\succeq0$ (or $\preceq 0$), but I have problems in proving more complicated cases.
Yes.
Let $U$ be the $nm×nm$ unitary matrix that swaps the order of the spaces. That is, $U(x\otimes y)= y\otimes x$ holds for every choice of vectors $x\in\mathbb{R}^n$ and $y\in \mathbb{R}^m$. Consider the mapping $\Psi$ defined as $\Psi(X) = UXU^*$ for every choice of $nm×nm$ matrix $X$. Then $$ \mathcal{I}\otimes\Phi = \Psi\circ (\Phi\otimes\mathcal{I})\circ \Psi^* $$ and thus $\mathcal{I}\otimes\Phi$ is the composition of positive maps.