For $\chi$ being an irreducible character of the symmetric group $S_n$ and being $M$ a complex $n\times n$-matrix, I would like to show

$$ \sum_{\sigma, \rho \in S_n} \overline{\chi(\sigma)} \chi(\rho) \prod_{j=1}^n M_{\sigma_j, \rho_j} = \frac{ n! }{\chi(e)} \sum_{\sigma \in S_n} \chi(\sigma) \prod_{j=1}^n M_{j, \sigma_j} , $$ where the right-hand-side is proportional to the immanant of $M$, $$ \text{imm}(M) = \sum_{\sigma \in S_n} \chi(\sigma) \prod_{j=1}^n M_{j, \sigma_j} $$ and $e$ is the identity element in $S_n$. This expression appears as the scalar product of many-body quantum states with exotic exchange symmetries.

For one-dimensional representations (i.e. the trivial constant or the alternating representation), we get the identities $$ \sum_{\sigma, \rho \in S_n} \prod_{j=1}^n M_{\sigma_j, \rho_j} = n! ~ \text{perm}(M) \\ \sum_{\sigma, \rho \in S_n} \text{sgn}(\sigma) \text{sgn}(\rho) \prod_{j=1}^n M_{\sigma_j, \rho_j} = n! ~ \text{det}(M) $$ for the permanent and the determinant (related to bosons and fermions), respectively, which are shown using $\chi(\sigma \rho)=\chi(\sigma) \chi(\rho)$ and $\chi(e)=1$ for one-dimensional $\chi$.