Problem
Let $V=\mathbb R^n$ be the space of column vectors, and $M$ a positive definite symmetric $n\times n$ real matrix. Suppose the matrix $A\in M_n(\mathbb R)$ satisfies $MAM^{-1}=A^t$. Show that there exists an invertible $P\in M_n(\mathbb R)$ such that $P^tMP=I_n$ and $P^{-1}AP$ is diagonal.
I know that if $M>0$ and $A$ is symmetric, then since there exists $S\in M_n(\mathbb R)$ such that $S^tMS=I_n$. Then not that $S^tAS$ is still symmetric; thus, there exists an orthogonal $Q$ such that $Q^tS^tASQ$ is diagonal and $Q^tS^tMSQ=Q^tQ=I_n$. Hence, $P=SQ$ is the desired matrix. However, I don't know how to show that $A$ is symmetric.
There is no need nor requirement that $A$ be symmetric.
The question can be answered, as user10354138 mentioned in the comments, by recognizing that $A$ is symmetric in the right inner product. Since such avenue requires the foresight to think in those terms, below is a straightforward approach.
Because $M$ is symmetric and positive definite, there exists $R$ such that $M=R^tR$. Let $S=R^{-1}$. We have $M=(SS^t)^{-1}$ and so $$ A^t=MAM^{-1}=S^{-t}S^{-1}ASS^t. $$ This can be rearranged to look like $$ (S^{-1}AS)^t=S^{-1}AS. $$ Hence $S^{-1}AS$ is symmetric and there exists orthogonal $Q$ with $Q^tS^{-1}ASQ$ diagonal. If we let $P=SQ$, then $P^{-1}AP$ is diagonal and $$ P^tMP=Q^tS^tMSQ=Q^tQ=I. $$