Solution
= Solution
For $h\in G$, right-multiplying by $D_2(h)^{-1}$ gives
$$
D_1(h)TD_2(h)^{-1}=\sum_{g\in G}D_1(hg)B D_2(g^{-1}h^{-1}).
$$
Make the bijective substitution $k=hg$, so $g^{-1}h^{-1}=k^{-1}$. The sum becomes $\sum_kD_1(k)BD_2(k^{-1})=T$. Hence
$$
\boxed{D_1(h)T=TD_2(h)\quad(h\in G).}
$$
Thus $T$ is an <intertwining operator> from the second <group representation> to the first. This is the <Schur averaging of rectangular matrices> construction.