If $M$ is a positive definite matrix, is $AMA^{-1}$ a positive definite matrix?
Notes
I am assuming that $A$ is a general matrix. If this property does not hold true for general matrices then what conditions on $A$ are needed? My intuition is that the symmetric part of $AMA^{-1}$ is positive definite as $M$ has the same eigenvalues as $AMA^{-1}$.
Work thus far
Let $y=A^{-1}z$ and $z$ be an appropriately sized vector \begin{align} z^T AMA^{-1}z&= (yA)^TA My \\ &= y^T A^TAMy \end{align} If we assume that $A$ is unitary then $A^T=A^{-1}$ and we have \begin{align} y^T A^TAMy &= y^T A^{-1}AMy\\ &=y^TMy \end{align} $y^TMy\geq 0$ for any $y$ so this implies $z^TAMA^{-1}z\geq 0$ if $A$ is unitary.
Is there any why to show the same holds true in for any $A$?
$AMA^{-1}$ has the same characteristic polynomial as $M$, but need not be symmetric (or hermitian in the complex case). The symmetric part need not be positive definite. For example, consider $$ A = \pmatrix{1 & 3\cr 1 & 2\cr},\ M = \pmatrix{1 & 0\cr 0 & 2\cr},\ A M A^{-1} = \pmatrix{4 & -3\cr 2 & -1}$$ Note that $$\pmatrix{0\cr 1}^T AMA^{-1} \pmatrix{0\cr 1} < 0$$