Is it possible that $P^t A P$ is digonal, but $P^{-1} A P$ isn't?

Question

I'm searching for a matrix $A$ $\in$ $M(n,\mathbb{R})$, such that there exists $P$ $\in$ $Gl(n,\mathbb{R})$ satisfying the following condition: $$P^t A P\ \mbox{ is diagonal, but } P^{-1}AP \ \mbox{isn't diagonal}. $$

Is it possible exists such matrices?

Can anyone help me?

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Counter-example using user357980's suggestion

$$A = \begin{bmatrix} 3/2& -1/2 \\ -1/2 &1/2 \end{bmatrix}, \quad P= \begin{bmatrix} 0 & 1 \\ 2 & 1 \end{bmatrix}. $$

Answer 1

Suggestion: Consider $n = 2$ (simplest case). Take a diagonalizable matrix $A$ that does not look diagonal and let $P$ be a matrix of eigenvectors so that $P^{-1}AP$ is diagonal. I would be surprised if $P^T A P$ was also diagonal.