Let $P$ and $Q$ be positive definite matrices. Consider the following matrix equation
$$\label{star}\tag{$\star$}
XPX^\top - P = -Q, \quad X\in\mathbb{R}^{n\times n}.
$$



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> **My question.** Is it true that any solution of \eqref{star} can be written as $X=(P-Q)^{1/2}TP^{-1/2}$ with $T$ being an arbitrary orthogonal matrix?
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Solving a "reversed" Stein equation
I am considering the previous version of your question which contained two parts.


Part 1) We assume $Q<P$ which means that $P-Q$ is a strictly positive matrix. Of course this condition is a necessary condition for the equation to have a solution. So we assume $P, Q,P-Q$ are positive invertible matrices.


First note that the equation has always a solution. Here is a proof:


WLOG we may assume that the positive matrix $P-Q$ is a diagonal matrix. Because every positive matrix is orthonormally diagonalizable. We consider the inner product whose tensor matrix is $P$. Now a solution to your matrix equation corresponds to a basis consistong of mutually orthogonal vectors whose norms are not necessarilly unit but are determined by the entries of the diagonal matrix $P-Q$.
Now one can show that if $X$ is a solution of the equation in your question then $A={(P-Q)}^{-1/2}XP^{1/2}$ satisfies $AA^T=Id$


Part 2) The answer is yes if we further assume that $PQ=QP$.


Note that the spectrum of the solution $X={(P-Q)}^{1/2}T P^{-1/2}$ is equal to the spectrum of $T{(P-Q)}^{1/2}P^{-1/2}$ because $AB$ and $BA$ have the same spectrum. Now the commutativity assumption $PQ=QP$ implies that $H={(P-Q)}^{1/2}P^{-1/2}$ is a positive matrix.


Now we use the following lemma:


**Lemma:** If $T$ is an orthonormal matrix and $H$ is a positive matrix Then $\lambda\_{min} (TH)= \lambda\_{min}(H)$ and $\lambda\_{Max} (TH)=\lambda\_{Max}(H)$.


**Proof:** It is sufficient to prove the following statement. Then the lemma is a consequence of a simple rescaling and consideration of the fact that "The spectrum of the inverse is equal to the inverse of the spectrum"


**Statement:** If all eigenvaluse of a positive matrix $H$ lie in the interior of the unit disc of the complex plane then the same is true for all eigenvalues of $TH$ where $T$ is an arbitrary orthonormal matrix:


**Proof of the Statement:** WLOG we may assume that $H$ is a diagonal matrix because every positive matrix is unitary equivalent to a diagonal matrix. For diagonal matrices the statement is obvious. Because if $THV=\lambda V$ for some $\lambda \in \mathbb{C}$ with $|\lambda|>1$ and $V \in \mathbb{C}^n$ then $|HV|>|V|$ which is impossible since all entries of diagonal matrix $H$ are positive element less than $1$.


The second part of the previous version of your question is a motivation to consider the following question in the context of complex $C^\*$ algebras. We keep the same notations $\lambda\_{min}$ and $\lambda\_{Max}$ for the case of $C^\*$ algebras.


**Question :** let $A$ be a simple $C^\*$ algebra and $u$ be a unitary element with the property that for every two positive element $a,b$ we have 


$$\lambda\_{min} (aub)=\lambda\_{min}(ab)\\ \lambda\_{Max}(aub)=\lambda\_{Max}(ab)$$


Does this imply that $u$ is an scalar element?


