It is easy to see that if $A$ and $B$ are two square matrices, then
$$(A\otimes B)^T = (A^T) \otimes (B^T)$$
(where $A\otimes B$ is the Kronecker product of $A$ and $B$ and $A^T$ is the transpose of $A$)
so, if $A$ and $B$ are symmetric, then $A\otimes B$ is also symmetric. Similarly, if $A$ and $B$ are skew-symmetric, then $A\otimes B$ is symmetric. My question is some kind of converse property: when is $A\otimes B$ symmetric?
In the case where $A$ and $B$ are non-symmetric $2\times 2$ matrices such that $A\otimes B$ is symmetric, I could prove that $A$ and $B$ are actually skew-symmetric (proof by writing $A$ and $B$ explicitly and considering a system of equations).
The notations sound cumbersome, so I am at a loss to generalize this observation. Any idea? Specifically, my question is:
If $A$ and $B$ are square matrices such that $A\otimes B$ symmetric, what can be said about $A$ and $B$?
The condition $A$ and $B$ square is used to rule out trivial examples where $A$ or $B$ is a non-square (hence non-symmetric) zero matrix.
Clearly $A\otimes B$ is symmetric when one of $A$ and $B$ is zero.
Suppose both $A$ and $B$ are nonzero. Then some $b_{rs}\ne0$. Since $A\otimes B$ is symmetric, we have $a_{ij}B=a_{ji}B^T$ for all $(i,j)$. Therefore $a_{ij}=ka_{ji}$ where $k=\frac{b_{sr}}{b_{rs}}$. In turn, $a_{ij}=ka_{ji}=k(ka_{ij})=k^2a_{ij}$ for all $(i,j)$. Since $A$ is nonzero, $k=\pm1$. Hence $A$ is either symmetric (when $k=1$) or skew-symmetric (when $k=-1$).
As $A\otimes B$ is similar to $B\otimes A$ via a simultaneous permuation of rows and columns, $B\otimes A$ is also symmetric. Therefore $B$ is also either symmetric or skew-symmetric.
It follows that $A$ and $B$ are either both symmetric or both skew-symmetric when none of them is zero.