Assume we have a matrix:
$$ (A^TB^{-1}A)^{-1}. $$
Under what circumstances can we simplify this to:
$$ C^TBC? $$
And what would $C$ be?
EDIT: Note that I am NOT assuming that $A$ is a square matrix. Only that $B$ is a square and invertible. In fact, let's assume $A$ is $n\times m$ with $n\neq m$, and $rank(A)=\min(n,m)$. Sorry for not stating this more explicitly earlier.
When you speak of inverse of a product, you can see $(AB)^{-1}=B^{-1} A^{-1}$. In your case you have $C = (A^T)^{-1}$.
If we should have $A^TB^{-1}A$ invertible then $$|A^TB^{-1}A|\ne 0$$also $B$ is invertible so $|B^{-1}|=\dfrac{1}{|B|}\ne 0$ and we have:$$|A^TB^{-1}A|=|A^T||B^{-1}||A|=|B^{-1}||A|^2\ne 0\to|A|^2\ne 0\to |A|\ne 0$$therefore $A$ is invertible as well. Now check that the following equalities are true:$$(A^TB^{-1}A)(A^{-1}B(A^T)^{-1})=A^TB^{-1}AA^{-1}B(A^T)^{-1}=A^TB^{-1}B(A^T)^{-1}=A^T(A^T)^{-1}=I$$also$$(A^T)^{-1}=(A^{-1})^{T}$$then by defining $C=(A^{-1})^T$ we have:$$(A^TB^{-1}A)^{-1}=C^TBC$$
If you multiply two matrices $A,B$ and $AB$ is invertible then both $A,B$ must be invertible. So if $A^TB^{-1}A$ is invertible then so are all of the terms appearing. Therefore we using that $(AB)^{-1}=B^{-1}A^{-1}$ and $(A^T)^{-1}=(A^{-1})^{T}$ we get that:
$(A^TB^{-1}A)^{-1}=A^{-1}B(A^{-1})^{T}$
So $C=(A^{-1})^T$
