Prove or disprove $ABA^{T} = AC$ implies $BA^{T} =C$ Suppose A is $m \times n$ non-zero matrix, $B$ is $n \times n$ of full rank and $C$ is $n \times m$ of full column rank. Can we prove or disprove that $ABA^{T} = AC$ implies that $C = BA^{T}$.
Note: Only $B$ is a square matrix. Rest are rectangular matrices with $m < n$.

$A=\begin{pmatrix}1 & 0\\
0 & 0
\end{pmatrix}$, $B=C=I_{2}$
$ABA^{T}=\begin{pmatrix}1 & 0\\
0 & 0
\end{pmatrix}=AC$ but $BA^{T}=\begin{pmatrix}1 & 0\\
0 & 0
\end{pmatrix}\neq C$
