Let $C_\mu$ be the size of the conjugacy class in $S_n$ of permutations whose cycletype is the partition $\mu\vdash n$. Let $\chi$ be the characters of the irreducible representations of $S_n$.

Let $\omega\vdash m$ and let $\theta\vdash(n+m)$. I am interested in the sum $$ \frac{1}{n!}\sum_{\mu\vdash n} C_\mu\chi_\lambda(\mu)\chi_\theta(\mu\cup\omega),$$ where $\mu\cup\omega$ is a partition of $n+m$ containing the parts of $\mu$ and of $\omega$. Numerics suggest that for most pairs $(\lambda,\theta)$ this sum is zero. In the simplest case of $\omega=0$, for example, the only non-vanishing pair is $\theta=\lambda$ (and the result is 1).

To illustrate, these are the values of the sum when $\omega=(3)$, with $n=4$ and $n+m=7$ ($\lambda$ is labelling the rows and $\theta$ the columns, both in lexicographic order) $$\left[ \begin {array}{ccccccccccccccc} 1&0&0&1&0&-1&0&0&1&0&0&0&0&0&0 \\ 0&1&0&0&0&0&0&-1&0&0&0&1&0&0&0 \\ 0&0&1&-1&0&0&0&0&0&0&-1&0&1&0&0 \\ 0&0&0&0&1&0&-1&0&0&0&0&0&0&1&0 \\ 0&0&0&0&0&0&0&0&1&-1&1&0&0&0&1\end {array} \right] $$

A possible solution would be to write $\chi_\theta(\mu\cup\omega)=\sum_{\rho\vdash n} a_\rho(\omega) \chi_\rho(\mu)$. Is there a known way to accomplish this decomposition? (I'm thinking something like the Murnaghan-Nakayama rule)