For a linear (super)-operator $\Psi : \mathbb{C}^{n\times n} \to \mathbb{C}^{m\times m} $, I am wondering whethe
$$ \text{Id}_{k} \otimes \Psi \text{ is positive for each } k\ge 1$$
is equivalent to $$ \Psi \otimes \text{Id}_{k} \text{ is positive for each } k\ge 1$$ In particular, I want to know whether positiveness of $\text{Id}_{k} \otimes \Psi$ is equivalent to positiveness of $\Psi \otimes \text{Id}_{k}$ or not.
Thanks in advance.
Late to the party, but tl;dr yes, the equivalence in question holds.
In the following, $\mathcal B(\cdot)$ denotes the $C^*$-algebra of linear and bounded operators on the respective Hilbert space. For further references, I recommend the book "Fundamentals of the Theory of Operator Algebras, Vol. I" by Kadison and Ringrose (1983), later referred to as [KR83]. Tensor products of Hilbert spaces and linear operators on such spaces are treated in Chapter 2.6 (p.125 ff.) quite thorougly.
The key for our main result will be the following isometric isomorphisms.
Proof. Consider the (obviously invertible) map $$ \forall_{h\in\mathcal H,g\in\mathcal G}\qquad\Phi_{\mathcal H}(h\otimes g):=g\otimes h $$ and its linear and continuous extension onto all of $\mathcal H\otimes\mathcal G$. Note that $\Phi_{\mathcal H}$ is isometric due to $$ \langle h_1\otimes g_1,h_2\otimes g_2\rangle_{\mathcal H\otimes\mathcal G}=\langle h_1,h_2\rangle_{\mathcal H}\langle g_1,g_2\rangle_{\mathcal G}=\langle g_1,g_2\rangle_{\mathcal G}\langle h_1,h_2\rangle_{\mathcal H}=\langle g_1\otimes h_1,g_2\otimes h_2\rangle_{\mathcal G\otimes\mathcal H}\tag{1} $$ which is the defining property of the scalar product on tensor products of Hilbert spaces, so $\mathcal H\otimes\mathcal G\simeq\mathcal G\otimes\mathcal H$. As $\Phi_{\mathcal H}\in\mathcal B(\mathcal H\otimes\mathcal G,\mathcal G\otimes\mathcal H)$ it admits a unique adjoint $\Phi_{\mathcal H}^\dagger\in\mathcal B(\mathcal G\otimes\mathcal H,\mathcal H\otimes\mathcal G)$. Again using (1), one quickly sees $\Phi_{\mathcal H}^\dagger=\Phi_{\mathcal H}^{-1}$. Therefore, the map $$ \Gamma_{\mathcal H}:\mathcal B(\mathcal H\otimes\mathcal G)\to \mathcal B(\mathcal G\otimes\mathcal H)\qquad A\mapsto\Phi_{\mathcal H}A\Phi_{\mathcal H}^\dagger $$ is bijective (with inverse $B\mapsto \Phi_{\mathcal H}^\dagger B\Phi_{\mathcal H}$) and, unsurprisingly, satisfies $$ \forall_{H\in\mathcal B(\mathcal H),G\in\mathcal B(\mathcal G)}\qquad\Gamma_{\mathcal H}(H\otimes G)\equiv G\otimes H \tag{2} $$ as another consequence of (1). Now $\Phi_{\mathcal H}$ being an isometric isomorphism implies $$ \|\Gamma_{\mathcal H}(A)\|=\sup_{x\in\mathcal G\otimes\mathcal H,\|x\|=1}\|\Phi_{\mathcal H}A\Phi_{\mathcal H}^\dagger x\|=\sup_{x\in\mathcal G\otimes\mathcal H,\|x\|=1}\|A\Phi_{\mathcal H}^\dagger x\|=\sup_{y\in\mathcal H\otimes \mathcal G,\|y\|=1}\|Ay\|=\|A\| $$ for all $A\in \mathcal B(\mathcal H\otimes\mathcal G)$ so, in total, $\mathcal B(\mathcal H\otimes\mathcal G)\simeq \mathcal B(\mathcal G\otimes\mathcal H)$.$\qquad\square$
Proof. Using the maps from the previous lemma (where now $\mathcal G=\mathbb C^k$), we need the following properties:
This is obvious due to the explicit form of $\Gamma_{\mathcal H}$ and the fact that $\Psi_{\mathcal H}$ is an isomorphism.
Use $\mathcal H\otimes\mathbb C^k\simeq\mathcal H\oplus\ldots\oplus \mathcal H$ ($k$ times; [KR83, Remark 2.6.8]) as then, elements from $\mathcal B(\mathcal H\otimes\mathbb C^k)$ can be identified with $k\times k$ matrices which have entries in $\mathcal B(\mathcal H)$ ([KR83, p.147 ff.]). Now let any $A\in\mathcal B(\mathcal H\otimes\mathbb C^k)$, so there exist $A_{ij}\in\mathcal B(\mathcal H)$ for $i,j=1,\ldots,k$ such that $A=\sum_{i,j=1}^n A_{ij}\otimes E_{ij}$ where $(E_{ij})_{i,j=1}^k$ is the standard basis of $\mathbb C^{k\times k}$. One gets $$ [\Gamma_{\mathcal K}\circ(\Psi\otimes\operatorname{id}_{k})](A)=\Gamma_{\mathcal K}\Big(\sum_{i,j=1}^k \Psi(A_{ij})\otimes E_{ij}\Big)\overset{(2)}=\sum_{i,j=1}^k E_{ij}\otimes\Psi(A_{ij})=(\operatorname{id}_{k}\otimes\Psi)\Big(\sum_{i,j=1}^k E_{ij}\otimes A_{ij}\Big)\overset{(2)}=[(\operatorname{id}_{k}\otimes\Psi)\circ\Gamma_{\mathcal H}]\Big(\sum_{i,j=1}^k A_{ij}\otimes E_{ij}\Big)=[(\operatorname{id}_{k}\otimes\Psi)\circ\Gamma_{\mathcal H}](A)\,. $$ With all of the above, getting the desired result is quite easy. Assume that $\Psi\otimes\operatorname{id}_{k}$ is positive. Now consider arbitrary $0\leq B\in\mathcal B(\mathbb C^k\otimes\mathcal H)$ so there exists unique $0\leq A\in\mathcal B(\mathcal H\otimes\mathbb C^k)$ with $\Gamma_{\mathcal H}(A)=B$ (as $\Phi$ is isomorphic). Then $$ (\operatorname{id}_{k}\otimes\Psi)(B)=[(\operatorname{id}_{k}\otimes\Psi)\circ\Gamma_{\mathcal H})](A)=\overbrace{[\Gamma_{\mathcal K}\circ\underbrace{(\Psi\otimes\operatorname{id}_{k})](A)}_{\geq 0}}^{\geq 0}\geq 0 $$ so $\operatorname{id}_{k}\otimes\Psi$ is positive as well. The converse is shown analogously. $\qquad\square$
Final Remark. At no point we explicitely used that we consider $\operatorname{id}_k$ so the statement still holds if we replace it by any linear map on $\mathbb C^k$.
Footnote (a): This means $A$ is positive semi-definite, i.e., $\langle x,Ax\rangle\geq 0$ for all $x\in\mathcal H\otimes\mathbb C^k$ as we are in a complex Hilbert space; in real Hilbert spaces we would have to additionally require $A$ being self-adjoint.