$A^TPA-P$ is negative definite, does it imply that $APA^T-P$ is negative definite

Question

If $P$ is positive definite, and if $A^TPA-P$ is negative definite, does it imply that $APA^T-P$ is negative definite?

Answer 1

No. Take this example $$P=\begin{pmatrix}1&0\\0&2\end{pmatrix},\quad A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$

Then $$A^TPA=\begin{pmatrix}0&0\\0&1\end{pmatrix}<P$$ but $$APA^T=\begin{pmatrix}2&0\\0&0\end{pmatrix}\not\le P$$