Consider real square matrices $X$ and $A$ of same size, where $A$ is known to be symmetric positive definite. I came across the matrix equation $XX^{\top} = AX^{\top}$, which solved for $X$ gives either $X = 0$, or $X=A$, or det$(X) = 0$ and det$(X-A)=0$. It is the third case ($X$ is neither equal to 0, nor equal to $A$) I seek to understand more.
Can something more be said about a singular matrix $X$ that solves det$(X-A) = 0$? I am hoping to use the additional information (symmetric positive definiteness) known about $A$, but not sure how to go about it.