I want to prove that every matrix is similar to its transpose. My lecturer gave us this exercise:
Let $\Bbb{F}$ be a field, $A\in M_{n\times n}$ and $A^t$ its transpose. We define $M,L=F^n$ to be $\Bbb{F}[x]$ modules, s.t for every $m\in M, f(x)m=f(A)m$ and for every $l\in L, f(x)l=f(A^t)l$. Prove that both modules are isomoprhic and conclude that $A\sim A^t$.
I proved that $M\cong L$ by proving that $\operatorname{SNF}(xI-A)=\operatorname{SNF}(xI-A^t)$, but I don't know how to conclude that the matrices are similar. Any help would be appreciated (it is important that we use the fact that those modules are isomorphic and not use other methods such as rational canonical form because we haven't learned those).
If $\varphi : M \to L$ is an isomorphism, $$(\forall m \in M) \quad \varphi(Am) = \varphi(xm) = x\varphi(m) = A^t \varphi(m).$$ So, if $P \in \textsf M_{n \times n}(\mathbb F)$ is the matrix such that $\varphi$ is left-multiplication by $P$ (every $\mathbb F$-linear map $\mathbb F^q \to \mathbb F^p$ is left-multiplication by some $p \times q$ matrix), the above equation tells us that $PA = A^tP$, hence $A = P^{-1}A^tP$.