The Schur orthogonality relation in coordinate/matrix element form reads $$\sum_{R\in G}^{|G|} \; \Gamma^{(\lambda)} (R)_{nm}^*\;\Gamma^{(\mu)} (R)_{n'm'} = \delta_{\lambda\mu} \delta_{nn'}\delta_{mm'} \frac{|G|}{l_\lambda}$$ where $\lambda,\mu$ lable the irreducible representations of group $G$. But there seems to be a case not specified clearly.
What happens when the two representations $\lambda$ and $\mu=\lambda'$ are equivalent but not identical to each other, i.e., related by a similarity transformation. What the relation would be in this case?
This is not an issue in the orthogonality relation in character form since Tr() is invariant under a similarity transformation. But for the matrix form above, the proof seems to be affected. For instance, when finding the normalization factor, one proof I saw uses (setting $\mu=\lambda$) $$c_{m'm}l_\lambda=\sum_{R}\sum_n \; \Gamma^{(\lambda)} (R^{-1})_{mn}\;\Gamma^{(\lambda)} (R)_{nm'} = \sum_{R} \; \Gamma^{(\lambda)} (R^{-1}R)_{mm'}.$$ But how about setting $\mu=\lambda'$? I feel it is not immediately true to have the second equality since $\Gamma^{(\lambda)}$ and $\Gamma^{(\lambda')}$ are not identical.