Find a $2\times2$ matrix $B$ such that $(A^TBA)^{-1}A^T=I$, where I is the identity matrix Problem
Let $A=\begin{pmatrix}
2 & 0 \\
1 & 2 
\end{pmatrix}$. Find a $2\times2$ matrix $B$ such that $(A^TBA)^{-1}A^T=I$, where $I$ is the identity matrix.
How does one go about solving this problem? I've been trying to wrap my head around what the question is exactly asking for. What hints can you drop so that I can solve this problem?
I've only recently learned about inverse matrices, and this is the first time I've seen this kind of question. What is the general way of going about solving these kinds of questions?

It might help to consider the analogous problem for scalars (i.e. numbers).
"If $a = 2$, find the scalar $b$ such that $(aba)^{-1}a = 1$". In this case, it would be clear how one should proceed: solve for $b$ and substitute $a$, in some order. For instance,
$$
(aba)^{-1}a = 1 \implies \frac{a}{aba} = 1 \implies \frac{1}{ba} = 1 
\\ \implies ba = 1 \implies b = a^{-1} \implies b = 2^{-1} = \frac 12.
$$
Of course it is more typical to substitute $a$ first, but with matrices it is often convenient to solve symbolically (as much as one can) before substituting.
Using the properties of the matrix inverse and transpose, try to do something similar with matrices $A$ and $B$.
