If $A$ is an $n \times m$ matrix with full row rank (in general not square) and $B$ is invertible, then does there exist $C, D, E$ such that


$$AXA^T = B \iff CX^{-1}D = E$$


for any symmetric positive semidefinite invertible matrix $X$?


If not, is there another way to transform $AXA^T = B$ into a matrix equation in terms of $X^{-1}$ with identical symmetric, positive definite solutions?



Are there matrices $C,D,E$ such that $AXA^T = B \iff CX^{-1}D = E?$
Converting previous comments into an answer, because I noticed they can fully answer the question.


First of all, one can change basis to assume $A = [I\,\, 0]$, and partition
$$
X = \begin{bmatrix}X\_{11} & X\_{12} \\ X\_{21} & X\_{22}\end{bmatrix}, \quad Y = X^{-1} = \begin{bmatrix}Y\_{11} & Y\_{12} \\ Y\_{21} & Y\_{22}\end{bmatrix},
$$
so the first equation becomes $X\_{11} = B$.


First of all note that the question makes sense: a condition in that form is not completely hopeless, because for a (non-posdef) matrix partitioned into four blocks of the same size for instance one has $X\_{11}=0 \iff Y\_{22}=0$, which are conditions in the form OP wants.


Using Schur complements,


$$B^{-1} = X\_{11}^{-1} = Y\_{11} - Y\_{12}Y\_{22}^{-1}Y\_{21}.$$


This latter condition is not an affine one like the second equation, and one can see indeed that it cannot converted to one, already in the case when all blocks have size $1$ (hence $A$ is $1\times 2$). In this case, for dimension reasons $C$ and $D^T$ must also be $1\times 2$, so that one gets one condition on $Y$. Then the relation that we obtain between the entries of $Y$, $$(y\_{11} - b^{-1})y\_{22} = y\_{12}y\_{21}, \tag{\*}$$ has degree 2, and cannot be reduced to to an affine one $$e = c\_{1}y\_{11}d\_{1} + c\_{1}y\_{12}d\_2 + c\_2y\_{21}d\_1 + c\_2 y\_{22}d\_2. \tag{\*\*}$$


(To justify this more formally, fix $y\_{11}$ and $y\_{22}$: then $(\*\*)$ is satisfied by $y\_{21},y\_{12}$ on a straight line, $(\*)$ by $y\_{21},y\_{12}$ on a hyperbola.)


