Find positive definite $X$ such that $AXA^T=\alpha X$. Given an invertible real matrix $A\in\mathbb{R}^n$, is it possible to obtain a (real) positive definite matrix $X$ (and scalar $\alpha>0$) such that
$$
AXA^T=\alpha X?
$$
Attempt: The previous equation can be written as $AXA^T-X +Q=0$ with $Q=(1-\alpha)X$. So, if $(1-\alpha)>0$ this looks like the Lyapunov equation here. However, the analytic solution to the Lypunov equation has the form
$$
X = \sum_{k=0}^\infty A^kQ(A^T)^k = (1-\alpha)\sum_{k=0}^\infty A^kX(A^T)^k
$$
So, even if such equation have a solution, I don't have a clue on how to obtain it from here.
Do you have other thoughts/suggestions?
 A: If the equation is solvable, by taking determinants on both sides, we must have $\alpha=\det(A)^2$. If this is the case, we may put $B=\det(A)^{-1}A$ and equation reduces to $BXB^T=X$.
It is solvable if and only if $B$ is similar to a real orthogonal matrix.
Suppose $BXB^T=X$. Then $(X^{-1/2}BX^{1/2})(X^{-1/2}BX^{1/2})^T=I$. Hence $X^{-1/2}BX^{1/2}$ is real orthogonal.
Conversely, suppose $B=PQP^{-1}$ for some real orthogonal matrix $Q$. Let $X=PYP^T$ (i.e. $Y=P^{-1}X(P^{-1})^T$). The equation then further reduces to $QY=YQ$. Hence any positive definite matrix $Y$ that commutes with $Q$ will give rise to a solution $X$. In particular, we can always take $Y=I$ to obtain a solution $X=PP^T$.
A: Consider $A = \pmatrix{2 & 1 \\ 0 & 1}$. Then you seek a matrix
$$
X = \pmatrix{a & b \\ c & d}
$$
and a constant $K$ such that
$$
\pmatrix{2 & 1 \\ 0 & 1}\pmatrix{a & b \\ c & d}\pmatrix{2 & 0 \\ 1 & 1} = K \pmatrix{a & b \\ c & d}
$$
i.e.,
$$
\pmatrix{2a+c & 2b+d \\ c & d}\pmatrix{2 & 0 \\ 1 & 1} = K \pmatrix{a & b \\ c & d}
$$
so
$$
\pmatrix{4a+2c+2b+d  & 2b+d \\ 2c+d & d} = K \pmatrix{a & b \\ c & d}
$$
Case 1: $d \ne 0$. Then $K = 1$, and we have
$$
\pmatrix{4a+2c+2b+d  & 2b+d \\ 2c+d & d} = \pmatrix{a & b \\ c & d}
$$
whence $2b+d = b$ so $b = -d$, and similarly $c = -d$; thus our equation becomes
$$
\pmatrix{4a+2(-d)+2(-d)+d  & -d \\ -d & d} = \pmatrix{a & -d \\ -d & d}\\
\pmatrix{4a-3d  & -d \\ -d & d} = \pmatrix{a & -d \\ -d & d}
$$
so $4a - 3d = a$, hence $a = d$, and we have
$$
X = \pmatrix{d & -d \\ -d & d}
$$
which is not positive definite, because its determinant is zero.
Case 2: $d = 0$. Then we want
$$
\pmatrix{4a+2c+2b & 2b \\ 2c & 0} = K \pmatrix{a & b \\ c & 0}
$$
For $X$ to have full rank, we need $c \ne 0$, so $K = 2$, and we can write
$$
\pmatrix{4a+2c+2b & 2b \\ 2c & 0} = \pmatrix{2a & 2b \\ 2c & 0}
$$
which reduces to $4a + 2c + 2b = 2a$, so $a + b + c = 0$. That gives
$$
X = \pmatrix{-(b+c) & b \\ c & 0}
$$
or, replacing $b$ and $c$ with their negatives,
$$
X = \pmatrix{(b+c) & -b \\ -c & 0}
$$
The determinant here is $bc$, so $b$ and $c$ must have the same sign for $X$ to be positive definite; the trace is $b+c$, so that sign must be "+". But then the quadratic formula gives the eigenvalues as
$$
x = \frac{(b+c) \pm \sqrt{ (b+c)^2 + 4bc}}{2}
$$
and because $(b+c)^2 + 4bc > (b+c)^2$, one of the two choices in the numerator leads to a negative eigenvalue, so $X$ is not positive definite.
In short: for the matrix $A$ given above, there is no solution $X$.
A: Using vectorization and Kronecker product properties this may be rewritten as
$$
(A \otimes A) \operatorname{vec} X = \alpha \operatorname{vec} X
$$
This is a regular eigenvalue problem for matrix $A \otimes A$. Moreover, eigenvalues of this matrix are $\{\lambda_i \lambda_j\}_{i,j=1,n}$ with corresponding eigenvectors $X$ being Kronecker products of eigenvectors $x_i \otimes x_j$. Here $\lambda_i, x_i$ are eigenvalues and eigenvectors of $A$.
Converting back from vectorized form to matrix form we obtain
$$
X_{(i,j)} = x_i x_j^\top\\
\alpha_{(i,j)} = \lambda_i \lambda_j
$$
There might be problems with this approach when $A$ has complex-valued eigenvalues or does not have full set of eigenvectors in case of eigenvalues with multiplicity greater than one.
As you see, matrix $X_{(i,j)}$ are rank-1 matrices, so they cannot be (strictly) positive definite. To obtain a positive definite matrix you need to linearly combine $n$ of them, all with the same value $\alpha$. This is rather rare case.
