I am wondering about solutions to the equation, with unknown $X$, $$ X^Tu - A^TXb = A^Tu. $$ where $X,A\in R^{n\times p}$, $A$ is full rank $\min(n,p)$, and the vectors $u, b\in R^n$ are nonzero.
This can be simplified slightly. If the SVD of $A$ is $A=PDQ^T$ with $D$ diagonal, it is equivalent to $$ QQ^TX^TPP^Tu - QDP^TXQQ^Tb = QDP^Tu, \qquad Y^Tv - D Yc = Dv $$ with $D$ diagonal and $Y=P^TXQ$, $c=Q^Tb$ and $v=P^Tu$.
Question: With $n,p\ge 2,$ does the first equation always have a solution $X$ in terms of $(y,A,b)$, or does the second equation always have a solution in terms of $(v,D,c)$?
I don't know the answers to your questions, but the equation is insoluble if and only if $u=Ab\ne0$.
If $u=Ab$, the equation becomes $(X^TA-A^TX)b = A^TAb$. Since $X^TA-A^TX$ is skew-symmetric, we obtain $0 = b^T(X^TA-A^TX)b = b^TA^TAb = \|Ab\|_2^2$. Therefore the equation is soluble only if $Ab=0$. If this is the case, $X=A$ is clearly a solution.
If $u\ne Ab$, we may try $X^T=A^TS$ for some symmetric matrix $S$. The equation then becomes $A^TSv=A^Tu$, where $v=u-Ab\ne0$. This is always soluble by taking, for instance, $$ S=\frac{uv^T+vu^T}{v^Tv}-\frac{(u^Tv)vv^T}{(v^Tv)^2}. $$