Can $AXA+BXB$ be simplified to $KXK$? $A, X$ and $B$ are real symmetric square matrices. Do we have a $K$ in terms of $A$ and $B$ such that the expression can be written as $KXK$? I used the "vec" trick to simplify the expression assuming that it can be done.
$$\left(A \otimes A\right)vec\left(X\right)+\left(B \otimes B\right)vec\left(X\right)=\left(K\otimes K\right)vec\left(X\right)\\A\otimes A+B\otimes B=K\otimes K$$
Is there a way to find $K$ from $K \otimes K$?
 A: It is not necessarily the case that $A \otimes A + B \otimes B$ can be written in the form $K \otimes K$. In fact, we find that there will exist such a $K$ if and only if $A,B$ are multiples of each other.
For example, taking
$$
A = \pmatrix{1&0\\0&2}, \quad B = \pmatrix{2&0\\0&1}
$$
leads to an $A \otimes B$ that cannot be written in this form.

If we know that $M = A \otimes A + B \otimes B$ can be written in the form $M = K \otimes K$, then we can find this $K$ as follows: suppose that $A,B$ have size $n$.
Let $\phi : \Bbb R^{n^4} \to \Bbb R^{n^4}$ denote the linear map (permutation) defined so that
$$
\phi(u \otimes v \otimes x \otimes y) = u \otimes x \otimes v \otimes y.
$$
We find that $M$ can be written in the form $M = K_1 \otimes K_2$ if and only if the matrix $P  = \operatorname{vec}^{-1}(\phi(\operatorname{vec}(M)))$ has rank $1$. It can be written in this form with $K_1 = K_2$ if and only if $P$ is also symmetric and positive semidefinite.
If $P$ is rank $1$, symmetric, and positive semidefinite, then $P$ can be written in the form
$$
P = vv^T
$$
for some vector $v$. If we take $K = \operatorname{vec}^{-1}(v)$, then we find that $M = K\otimes K$ as desired.

You might find it preferable to note that
$$
\operatorname{vec}^{-1}(\phi(\operatorname{vec}(A \otimes B))) = \operatorname{vec}(B) \operatorname{vec}(A)^T
$$
and use $A \otimes B \mapsto \operatorname{vec}(B) \operatorname{vec}(A)^T$  instead of my definition of $\phi$.
