Schur's lemma concerns 2 arbitrary irreducible representations of a finite group, say $D_{(1)}(A_k), D_{(2)}(A_k)$ on a group $\{A_1, A_2, \dots\}$. It considers the consequences of the existence of a matrix $M$ satisfying
$$MD_{(1)} = D_{(2)}{M}$$
In a proof I've seen it's assumed that $D_{(1)}$ and $D_{(2)}$ are already in unitary form. I want to make sure I understand why. I do understand that any representation can be put into unitary form. That being said, is it because if the matrices $U_{(1)}, U_{(2)}$ bring $D_{(1)}$ and $D_{(2)}$ respectively into unitary form, then we can say that
$$ \begin{split} U_{(2)}MD_{(1)}U_{(1)-1} &= U_{(2)}D_{(2)}MU_{(1)}^{-1} \\ U_{(2)}MU_{(1)}^{-1}U_{(1)}D_{(1)}U_{(1)}^{-1} &= U_{(2)}D_{(2)}U_{(2)}^{-1}U_{(2)}MU_{(1)}^{-1} \\ (U_{(2)}MU_{(1)}^{-1})(U_{(1)}D^{(1)}U_{(1)}^{-1}) &= (U_{(2)}D_{(2)}U_{(2)}^{-1})(U_{(2)}MU_{(1)}^{-1}) \\ \hat{M}\hat{D}_{(1)} &= \hat{D}_{(2)}\hat{M} \end{split} $$
where $\hat{D}_{(1)}, \hat{D}_{(2)}$ are both unitary? Or is there a more direct argument which I'm missing?