Sum of product of characters of $S_{n+m}$ over $S_n$ I asked this in MSE first, here, but it is not getting any attention.
Let $S_n$ be the group of permutations acting on the set $\{1,...,n\}$.
Given $R_1,R_2$ two irreps of $S_n$ with characters $\chi_{R_i}$, I know that
$$ \sum_{a\in S_n}\chi_{R_1}(a)\chi_{R_2}(ab)=n!\frac{\chi_{R_1}(b)}{\chi_{R_1}(1)}\delta_{R_1R_2},$$
when $b\in S_n$.
What I want to know is what is the result of
$$ \sum_{a\in S_n}\chi_{R_1}(a)\chi_{R_2}(ab),$$
where now $R_1$ is an irrep of $S_n$ but $R_2$ is an irrep of $S_{n+m}$, with $b\in S_{n+m}$ and $a$ having $\{n+1,...,n+m\}$ as fixed points. Can this be expressed in terms of the representation induced from $R_1$, for example?
 A: I would rather write $\chi_{R_1}(a)=\chi_{R_1^*}(a^{-1})$.  This allows us to see the sum
$$\frac{1}{n!}\sum_{a\in S_n}\chi_{R_1^*}(a^{-1})\chi_{R_2}(ba)$$ as the trace on $\mathrm{Hom}_{\mathbb{C}}(R_1,R_2)$ of the map $f\mapsto \frac{1}{n!}\sum_{a\in S_n}bafa^{-1}=b\frac{1}{n!}\sum_{a\in S_n}afa^{-1}$.  If $b=1$, this is projection to the space of invariants, so its trace is the dimension of this space $\delta_{R_1,R_2}$.  If $b\neq 1$, then it's projection to invariants, followed by composition with $b$.  If you choose a basis consisting of $b$ (which is sent to $\chi_{R_1}(b)/\chi_{R_1}(1)$ times itself), and then a basis of the other isotypic components in $\mathrm{Hom}_{\mathbb{C}}(R_1,R_2)$ (which are all killed), you can see this gives the factor of $\chi_{R_1}(b)/\chi_{R_1}(1)$ (incidentally, this seems to be a restatement of the proof that this ratio is an algebraic integer).
For the generalization, we still have that $$\frac{1}{n!}\sum_{a\in S_n}\chi_{R_1^*}(a^{-1})\chi_{R_2}(ba)$$ as the trace on $\mathrm{Hom}_{\mathbb{C}}(R_1,R_2)$ of the map $f\mapsto \frac{1}{n!}\sum_{a\in S_n}bafa^{-1}=b\frac{1}{n!}\sum_{a\in S_n}afa^{-1}$. However, this projection might have a much larger image, since there might be many homomorphisms in $\mathrm{Hom}_{S_n}(R_1,R_2)$, since $R_2$ isn't irreducible as an $S_n$-module. My impulse is that there's no easy expression for this trace, but I can't say I'm sure.
