Suppose $U$ is an orthogonal matrix.
Show that $T : M_{22} → M_{22}$ defined by $T(A) = UAU^{−1}$ is an isometry. I know that a linear operator is an isometry if and only if it sends some orthonormal basis to an orthonormal basis. But I have no idea how to start this.
I'm assuming you're using the standard inner product to define a norm, $\langle A,B\rangle = \text{tr}(A^TB)$. In this case, $\|A\| = \text{tr}(A^TA)$, and we have $$ \text{tr}((UAU^T)^T(UAU^T)) = \text{tr}(UA^TU^TUAU^T) = \text{tr}(UA^TAU^T) = \text{tr}(U^TUA^TA)=\text{tr}(A^TA)$$ where we used the cyclic property of the trace and the fact that for orthogonal $U$ we have $U^{-1}=U^T$, showing that your map is indeed an isometry.