We say that two matrices $A,\,B\in M_n(R)$ are similar if there is some invertible matrix $P$ such that $P^{-1}AP=B$. Now, if $R$ was a field (or certainly an algebraically closed field) then it is straightforward to show $A$ and $A^T$ are similar. Simply use the Jordan form.
I am wondering if this result also holds true over more general rings, say a PID.
As a starting position I was thinking of looking over $\mathbb{Z}$ and perhaps using the Smith Normal Form in some way.
No, it's not!
The matrix $A=\pmatrix{8&2\\0&1}$ is not similar over $\mathbb Z$ to its transpose.
For another example, the matrix $A = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix}$ is not similar over $\mathbb{Z}$ to its transpose.
(I've just checked this, as I wondered whether assuming unitriangularity would help.)
There is also a counterexample over the ring $\mathbb{Q}\left[x\right]$ instead of $\mathbb{Z}$: The matrix $A = \begin{pmatrix} 1 & x & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$ is not similar over $\mathbb{Q}\left[x\right]$ to its transpose. (This shows that requiring characteristic $0$ does not help.)
