Yeah. Sorry, I just noticed that my way of asking is finding "a" $M$. Then your answer fully solves my problem and I have accepted the answer. The problem in the background is to find all such $M$. If you have any ideas welcome to share.

This answer finds all such $M,\alpha$ when $M$ and $H$ can be mutually diagonalizable. But for the scenario where $M$ is not a diagonal matrix after $H$ has been diagonalized, there may be choices of $M$ such that $A$ is stable. Nevertheless, @iosif-pinelis solves 99% of my question. Thank you!

This is really a nice answer. But this sentence may contain a typo: "So, $\lambda$ is a nonzero eigenvalue of $A$ iff $d(\lambda)=1$" should be "$d(\lambda)=0$". .