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