Let $A\in\mathbb{R}^{n\times n}$ be a Hurwitz stable matrix (i.e. all eigenvalues of $A$ have negative real part). Let $\succeq$ denote the standard partial order in the cone of positive semidefinite matrices, and consider the following iteration, for $k\in\mathbb{Z}$, $k\ge 0$,
$$\tag{1}\label{eq:1}
X_{k+1} = \frac{P_k^{1/2}X_k P_k^{1/2}}{\mathrm{tr}(P_k^{1/2}X_k P_k^{1/2})}, \ \  \ X_0\succeq 0,\ \mathrm{tr}(X_0)=1,
$$
where $P_k\succeq 0$ is the solution of $AP_k+P_kA^\top =-X_k$, and $\cdot^{1/2}$ denotes the principal matrix square root.

First, note that a fixed point of \eqref{eq:1} always exists by Brouwer's fixed point theorem (indeed, \eqref{eq:1} is continuous and maps the compact set of unit trace positive semidefinite matrices to itself). However, the fixed point is, in general, not unique.

Second, if $A+A^\top \prec 0$ then there exists a fixed point $\bar{X}\succeq 0$ of \eqref{eq:1} such that the solution $\bar{P}\succ 0$, where  $\bar{P}$ is the solution of $A\bar{P}+\bar{P}A^\top =-\bar{X}$ (to see this pick for instance $\bar{P}=-\frac{1}{2\mathrm{tr}(A)}I$).

> **Hence, my question:** Does there always (i.e., for any Hurwitz stable $A$) exist a fixed point $\bar{X}\succeq 0$ of \eqref{eq:1} such that $\bar{P}\succ 0$, where  $A\bar{P}+\bar{P}A^\top =-\bar{X}$?