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.
 A: Case 1: Square A
Claim: If A is square, then A is non-singular.
Say A is square and singular. Then $\exists M$ (of full rank) such that $MA$ does not have full rank.
Then, $A^\top MA$ does not have full rank. Therefore $A^\top MA$ is non-invertible (see here), leading to a contradiction. Thus A is non-singular.
Case 2: If $A$ is not square, and $A\in\mathbb R^{m\times n}$,($m>n$).
Claim: $\exists M$ such that $A^\top MA$ is not invertible.
Construct M such that $MA$ gives at least 1 vector in the null space of $A^\top$. Therefore $A^\top MA$ has rank less than n for such M, and is therefore non-invertible.
Illustration: if $A = \begin{bmatrix}1\\ 2\end{bmatrix}$,$A^\top A = \begin{bmatrix}1& 2\end{bmatrix} \begin{bmatrix}1\\ 2\end{bmatrix} = 5$. But since we know $v = \begin{bmatrix}-2& 1\end{bmatrix}$ is $\perp$ to $A$, we can construct $M$ such that $MA = v$. For example, $M = \begin{bmatrix}1&-1.5\\ 1&0\end{bmatrix}$ gives us $MA = \begin{bmatrix}-2\\ 1\end{bmatrix}$ and therefore $A^\top MA = \begin{bmatrix}0\end{bmatrix}$
Case 3: If $A$ is not square, and $A\in\mathbb R^{m\times n}$,($n>m$)
Claim: $\forall M$, $\quad A^\top MA$ is not invertible.
Note that $\operatorname {Rank}(A^\top MA)\leq \operatorname {Rank}(A) = m$, even though $A^\top MA \in\mathbb R^{n\times n}$. Therefore it is not full rank and not invertible
NOTE: I had earlier assumed a counterexample exists based on another answer,
A: Yes. In fact:
Claim: If $A$ is an $m \times n$ matrix such that for every invertible $m \times m$ matrix $M$, $A' M A$ is invertible, then $A$ is square and invertible.
This includes $M=I$, so $A'A$ is invertible, an $n \times n$ matrix. We have
$$
  n = \operatorname{rank} A'A \leq \operatorname{rank} A \leq m,
$$
so $n \leq m$, $A$ is "tall". We claim that in fact $n=m$.
Since $A$ has rank $n$, the columns of $A$ are independent. Call their span $U$. If $n < m$, then the orthogonal complement $U^\perp$ is nonzero.
There is an invertible matrix $M$ that moves at least one column of $A$ into $U^\perp$. Explicitly, let the columns of $A$ be $c_1,\dotsc,c_n$, and let $v_1$ be a nonzero vector in $U^\perp$. We can choose any invertible map $M$ that has $M c_1 = v_1$. (For the argument it doesn't matter what happens to the rest of the $c_i$. There's a lot of freedom.)
But then $A' M A$ has a zero column, because every row of $A'$ is one of the $c_i$'s (transposed), and multiplying that row $c_i$ with the column $v_1$ of $MA$ gives us a value of $c_i \cdot v_1 = 0$. So all the entries in that column are zero. Therefore $A' M A$ isn't invertible after all.
This contradiction shows $n=m$, $A$ is square.
Finally (as noted in comments to the other answer) now $A$ must be invertible: since $A'A$ is invertible, $\det(A'A) \neq 0$, but $\det(A'A) = \det(A)^2$. This finishes the proof of the claim.
Now if $A$ is square and invertible, then for any invertible $M$,
$$
  (A' M A)^{-1} A' M = A^{-1} M^{-1} A'^{-1} A' M = A^{-1} = (A'A)^{-1} A',
$$
the property you were looking for.
A: 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 $A$.
