Is it possible that $A \neq B$ and $AT^{-1}(A)=BT^{-1}(B)$, where $T(\cdot)$ is linear?

Question

Let $A,B \in \mathbb{R}^{n \times n}$ and $T:\mathbb{R}^{n \times n} \rightarrow \mathbb{R}^{n \times n}$ be a linear map. Suppose that $$AT^{-1}(A)=BT^{-1}(B)\,,$$ where $T^{-1}(A)$ is the matrix inverse of $T(A).$ Is it possible that $A\neq B$?

Answer 1

Yes, by linearity $B=-A$ also works.

Answer 2

Let $T:\mathbb{R}^{n \times n} \rightarrow \mathbb{R}^{n \times n}$ be defined by $T(A)=A^t$. Then $T^2=id$, hence $T^{-1}=T$.

Thus $AT^{-1}(A)=BT^{-1}(B)$ means $AA^t=BB^t$.

It is your turn to find an example that $AA^t=BB^t$ does not imply $A=B$