Let $G = GL(n, \mathbb{C})$ and $M = \{Q \in GL(n, \mathbb{C}) | Q^t = Q\}$. Let $P\in M$ be non-diagonalizable. Question: For any $A \in G$, can we say that $APA^t$ is also non-diagonalizable? Or there's no definite conclusion?
p.s. For simplicity, take $P=\left[\begin{array}{cc} 2i & 1 \\ 1 & 0\end{array}\right]$.