If $V$ and $W$ are two finite dimensional vector spaces and $\operatorname T$ and $\operatorname S$ be two invertible linear maps from $V$ to $W$, where $\operatorname T$ and $\operatorname S$ have same matrix representation wrt different ordered bases of $V$ and $W$.
Show that there exists two invertible linear maps $\operatorname{P}\colon V\longrightarrow V$ and $\operatorname{Q}\colon W\longrightarrow W$ such that $\operatorname T = \operatorname{Q^{-1}}\operatorname{S}\operatorname{P}$.
Actually, I was trying out by considering identity maps from $V$ onto $V$ wrt ordered bases $V_1$ and $V_2$ and the same for $W_1$ and $W_2$ but couldn't advance more.
Obviously firstly, I have to reach $V_2$ from $V_1$, then $W_2$ from $V_2$ and then $W_1$ from $W_2$ to make the composition of the maps equal to $\operatorname T$, i.e. from $V_1$ to $W_1$ .
So, you have two bases $B$ and $B^\star$ of $V$ and two bases $C$ and $C^\star$ of $W$ and you are assuming that the matrix of $T$ with respect to the basis $B$ and $C$ is equal to the matrix of $S$ with respect to the basis $B^\star$ and $C^\star$. Let $M$ be this matrix. Let $P\colon V\longrightarrow V$ be the linear endomorphism of $V$ which maps the $k$th vector of $B$ into the $k$th vector of $B^\star$. And let $Q\colon W\longrightarrow W$ be the linear endomorphism of $W$ which maps the $k$th vector of $C$ into the $k$th vector of $C^\star$. Then $T=Q^{-1}SP$ since the matrix of both linear maps with respect to the bases $B^\star$ and $C^\star$ are equal to $M$.