Kronecker product of $A$ and $B$ is similar to the Kronecker product of $B$ and $A$ Given two matrices $A$ and $B$. How would one prove that the Kronecker product of $A$ and $B$ is similar to the Kronecker product of $B$ and $A$?
 A: If $A$ and $B$ are square, then $A\otimes B = P^{-1}(B\otimes A)P$ where $P$ is a permutation matrix that you can write down explicitly; it depends only on the dimensions of $A$ and $B$.
If $A$ and $B$ are not square, then $A\otimes B$ and $B\otimes A$ are not necessarily similar even if they happen to be square. For example, if
$$ A=\begin{pmatrix}0\\1\end{pmatrix} \qquad
B = \begin{pmatrix}0&0&1&0\\0&0&0&1\end{pmatrix}$$
then $A\otimes B$ and $B\otimes A$ are not similar (they have different characteristic polynomials).
A: This is most easily seen in terms of coordinate-free linear algebra.
If $f : V \to W$ and $g : V' \to W'$ are linear maps and $S_{V,V'} : V \otimes V' \to V' \otimes V$ denotes the symmetry isomorphism, the diagram
$$\begin{array}{c} V \otimes V' & \xrightarrow{f \otimes g} & W \otimes W' \\ S_{V,V'} \downarrow ~ && ~ \downarrow S_{W,W'} \\ V' \otimes V & \xrightarrow{g \otimes f} & W' \otimes W \end{array}$$
commutes, since $v \otimes v'$ gets maped to $g(v') \otimes f(v) \in W' \otimes W$ by both compositions. If $W=V$ and $W'=V'$, it follows that $f \otimes g = S_{V,V'}^{-1} \circ (g \otimes f) \circ S_{V,V'}$, so that $f \otimes g,\,g \otimes f  \in \mathrm{End}(V \otimes V')$ are similar.
