Solving a "reversed" Stein equation 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?

 A: 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?
