Let $X \in \mathbb{R}^{n\times p}$ and $A \in \mathbb{R}^{p\times q}$ such that $\mathrm{rank}(XA) = \mathrm{rank}(X)$. I would like to prove that then there exists a matrix $B \in \mathbb{R}^{q \times p}$ such that $XAB = X$.
If $A$ is square and invertible, clearly one can take $B = A^{-1}$. If $A$ has linearly independent rows, one can take $B$ to be the Moore-Penrose inverse of $A$. Does $B$ exist for general $A$? If the statement is wrong in general, does it hold for $q = \mathrm{rank}(X)$?
This works over any field. As $\operatorname{range}(XA)\subseteq\operatorname{range}(X)$ and $\operatorname{rank}(XA)=\operatorname{rank}(X)$, we have $V:=\operatorname{range}(XA)=\operatorname{range}(X)$. Let $Y$ be any matrix whose columns form a basis of $V$. Then $X=YP$ and $XA=YQ$ for some matrices $P$ and $Q$. Since $YQ\,(=XA)$ has the same rank as $Y$ and $Y$ has full column rank, $Q$ must have full row rank. Therefore $Q$ has a right inverse $R$. Let $B=RP$. Then $XAB=YQRP=YP=X$.
This means $XAB=X$ is a consistent system of equations. So, if the underlying field is $\mathbb R$ or $\mathbb C$, $B=(XA)^+X$ is a solution to $XAB=X$.