Does $(A'MA)^{-1}A'M=(A'A)^{-1}A'$ for all nonsingular $M$ imply $A$ is nonsingular? Is the following statement true?
Suppose $A$ is a real matrix and $A'A$ is nonsingular. Suppose that for every nonsingular real matrix $M$, $(A'MA)^{-1}A'M=(A'A)^{-1}A'$, then $A$ is square and nonsingular.

I eventually came up with an alternative proof to the ones provided. Let $A$ be $n\times m$.  We know $A'A$ is a nonsingular $m\times m$ matrix, so (as pointed out in other answers) $m=rank(A'A)\leq rank(A)\leq n$. So if $m<n$ then $A'$ has a non-empty null space. Let $v$ be some non-zero element of the null space i.e., $A'v=0$, then $(A'A)^{-1}A'v=0$ and so for every non-singular $M$ $(A'MA)^{-1}A'Mv=0$ and thus $A'Mv=0$. Thus if $G$ and $H$ are non-singular $m\times m$ matrices $(G+H)v$ is in the null space of $A'$. Now, since $v$ is non-zero then for any vector $l$ there is a matrix $K$ so that $Kv=l$ (just set the $k^{th}$ row of $K$ to equal $l_k v'/(v'v)$ where $l_k$ is the $k^{th}$ entry of $l$. Any $m\times m$ matrix can be written as a sum of two non-singular matrices, so it follows that $l$ is also in the null space. But then every vector is in the null space of $A'$ which means $A'$ is zero and we have a contradiction. So $A$ is square and (as has already been pointed out) therefore nonsingular.
Edit: wrote $B$ instead of $A$.
