Kroenecker product When taking the kroenecker product of two matrices it should hold that $A\otimes B$ is unitarily equivalent to $B \otimes A$ but I can't figure out what this unitary matrix would look like. 
 A: Not only are they unitarily equivalent, they are Permutation similar. There exists a permutation matrix $P$ such that 
$$P (A\otimes B) P^\top = B \otimes A$$
A: Note that $A \otimes B$ and $B \otimes A$ can only be guaranteed to be unitarily similar if both $A$ and $B$ are square.
So, suppose $A$ is $n \times n$ and $B$ is $m \times m$. It suffices to build a matrix $P$ such that $P(x \otimes y) = y \otimes x$ for all $x \in \Bbb C^n, y \in \Bbb C^m$.  In this case, we find that for all such $x,y$ we have
$$
P (A \otimes B)P^{-1}(y \otimes x) = P(A \otimes B)(x \otimes y) = P(Ax \otimes By) = By \otimes Ax
$$
so that $P (A \otimes B)P^{-1} = B \otimes A$.
We can build $P$ explicitly as follows: Let $e_1,\dots,e_n$ denote the canonical basis of $\Bbb C^n$.  We can build the necessary permutation matrix $P$ by taking as the columns of $P$ the vectors $e_i \otimes e_j$ with $e_i \in \Bbb C^m, e_j \in \Bbb C^n$, and the tuples $(j,i)$ taken in lexicographical order.
In the case where $m=n$ we find that $P^* = P$, which is to say that $P$ is Hermitian (symmetric) in addition to being a permutation matrix.
As an example: for the case of $n=3,m=2$, we find that
$$
P = [e_1^{(2)} \otimes e_1^{(3)}, e_2^{(2)} \otimes e_1^{(3)},\dots]
= 
\left[\begin{array}{cc|cc|cc}
1&0&0&0&0&0\\
0&0&1&0&0&0\\
0&0&0&0&1&0\\
\hline
0&1&0&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&0&1
\end{array}\right]
$$
