Prove or disprove $ABA^{T} = AC$ implies $BA^{T} =C$

Question

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 $m < n$.

Answer 1

$A=\begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix}$, $B=C=I_{2}$

$ABA^{T}=\begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix}=AC$ but $BA^{T}=\begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix}\neq C$

Answer 2

Set $A=\begin{pmatrix} -1 & 1\end{pmatrix}$, $B=\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$, $C=\begin{pmatrix} 1 \\ 3\end{pmatrix}$.

Then $ABA^T=AA^T=2=AC$ but $BA^T=A^T\neq C$.