Suppose $A$ is an $nm \times nm$ block matrix over some field, with $n^2$ blocks each of size $m \times m$. Suppose further that the blocks commute. Let $A^{BT}$ denote the "block transpose" of $A$, namely the $nm \times nm$ block matrix whose $(j,i)$ block is the $(i,j)$ block of $A$. That is, the positions of the blocks have been transposed, but not the blocks themselves.
Is $A$ similar to $A^{BT}$?
If $m=1$, the answer is yes. I can show that $A$ and $A^{BT}$ have the same minimal and characteristic polynomials in general.
For example, suppose $A, B, C, D$ are commuting $2 \times 2$ matrices. Is it necessarily the case that
\begin{align*}
\begin{pmatrix}
A & B \\
C & D
\end{pmatrix}
\sim
\begin{pmatrix}
A & C \\
B & D
\end{pmatrix}?
\end{align*}
Best Answer
When $m\ge4$, the answer is "no". Consider $m=2k$ and $A,B,C,D\in M_{2k}(\mathbb F)$ such that $$ A=\pmatrix{0&X\\ 0&0}, \ B=\pmatrix{0&Y\\ 0&0}, \ C=\pmatrix{0&Z\\ 0&0}, \ D=\pmatrix{0&W\\ 0&0} $$ for some $X,Y,Z,W\in M_k(\mathbb F)$. Clearly $A,B,C,D$ commute, because the product of any two of them is zero. However, when $k\ge2$, we have $$ \text{rank of }\pmatrix{A&B\\ C&D}=\text{rank of }\pmatrix{X&Y\\ Z&W}\ne\text{rank of }\pmatrix{X&Z\\ Y&W}=\text{rank of }\pmatrix{A&C\\ B&D} $$ in general. Therefore the matrix on the left is not always similar to its blockwise transpose. E.g. let $k=2$ and $\{X,Z,Y,W\}$ be the standard basis of $M_k(\mathbb F)$. Then $$ \text{rank of }\left(\begin{array}{cc|cc}1&0&0&1\\ 0&0&0&0\\ \hline0&0&0&0\\ 1&0&0&1\end{array}\right) =2\ne4= \text{rank of }\left(\begin{array}{cc|cc}1&0&0&0\\ 0&0&1&0\\ \hline0&1&0&0\\ 0&0&0&1\end{array}\right). $$