We know that
$\int_{U(N)} \chi_{R}(A\Omega B\Omega^{\dagger}) [d\Omega ] = \chi_R(A)\chi_R(B)/\chi_R(1)$
where $[d\Omega ]$ is the Haar measure on the unitary group $U(N)$, $A, B$ are hermitian $N\times N$ matrices, $\Omega \in U(N)$ is an $N\times N$ unitary matrix and $\chi_R$ is an irreducible character of $GL(N, \mathbb{C})$ in the irreducible representation $R$. To derive this result, we can write
$\chi_R(A\Omega B\Omega^{\dagger}) = Tr(A^{(R)}\Omega^{(R)}B^{(R)}(\Omega^{\dagger})^{(R)}) = A^{(R)}_{ij}\Omega^{(R)}_{jk}B^{(R)}_{kl}(\Omega^{\dagger})^{(R)}_{li}$,
where $M^{(R)} = R(M)$ is a matrix in the representation $R$ and summation over repeated indices is implied. Then, we can use that
$\int_{U(N)} [d\Omega]\Omega^{(R)}_{ij}(\Omega^\dagger)^{(R')}_{kl} = \delta_{RR'}\delta_{ik}\delta_{jl}/d_{R}$,
where $d_{R}$ is the dimension of the representation $R$ and the result follows. Now, I'm trying to compute the following integral
$\int_{U(N)}[d\Omega]\chi_{R}(A\Omega B\Omega^{\dagger})\chi_{R'}(A^2\Omega B^2\Omega^{\dagger})$,
where $R, R'$ are different representations. In trying to use the same approach, I stumbled in the integral
$\int_{U(N)}[d\Omega]\Omega^{(R)}_{ij}(\Omega^\dagger)^{(R')}_{kl}\Omega^{(R)}_{mn}(\Omega^\dagger)^{(R')}_{rs}$.
If $R, R'$ are both the fundamental representation, this integral can be calculated using Weingarten calculus. Is there a way to compute such an integral for abitrary, possibly different representations $R$ and $R'$? If not, are there any methods to calculate the integral of products of characters of $GL(N, \mathbb{C})$ like the one I showed?
