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 $\chi$. 



Sum identities with immanants
I hope this is not homework.


\begin{eqnarray\*}
 \text{lhs}&=&\sum\_{\sigma,\rho} \overline{\chi(\sigma)}\chi(\rho)\left(\prod\_{i=1}^nM\_{\sigma(i),\rho(i)}\right)\\
 &=&\sum\_{\sigma,\rho} \overline{\chi(\sigma)}\chi(\rho)\left(\prod\_{i=1}^nM\_{i,\rho\sigma^{-1}(i)}\right)\\
 &=&\sum\_{\tau}\left(\sum\_{\sigma}\overline{\chi(\sigma)}\chi(\tau\sigma) \right)\prod\_{i=1}^n M\_{i,\tau(i)}\\
 &=&\sum\_{\tau}\frac{n!}{\chi(e)} \chi(\tau)\prod\_{i=1}^nM\_{i,\tau(i)}.
\end{eqnarray\*}


