Hyperoctahedral group acting on a special permutation Let $[n]=\{1,...,n\}$ and  $[\hat n]=\{\hat 1,...,\hat n\}$. Realize the hyperoctahedral group $H_n$ as the centralizer of the permutation $(1\hat 1)\cdots (n \hat n)$. It has $2^n n!$ elements.
Let $\pi$ be the special permutation $\pi=(12\cdots n)(\hat 1)\cdots(\hat n)$, i.e. the elements of $[n]$ are arranged in a cycle while the elements of $[\hat n]$ are all fixed points.
The question I would like to answer is this: How many elements $h\in H_n$ are there such that the product $h\pi$ has cycle type $\mu\vdash 2n$?
For example, if $\mu=1^{2n}$ then $h$ would have to be $\pi^{-1}$; since $\pi\notin H_n$, the solution is zero.
Numerics suggest that for $\mu=(2n)$ the solution is $2^{n-1}(n-1)!$
 A: It sounds to me like you are trying to compute zonal spherical functions by projecting the corresponding irreducible characters, that is,
$$ \omega^\lambda((1,3,\cdots,2n-1)) = \frac{1}{|H_n|}\sum_{h \in H_n} \chi_{2\lambda}((1,3,\cdots,2n-1)h),$$
where $H_n \leq S_{2n}$ is the hyperoctahedral group of order $2^nn!$. This formula is typically too involved to yield any sort of nice expression for the zonal spherical functions. In "Harmonic Analysis on Finite Groups" by Ceccherini-Silberstein, Scarabotti, & Tolli it is shown how to compute the zonal spherical functions combinatorially (the computations get complicated in a hurry), but in your case, you'd end up re-deriving the following.
The zonal spherical functions are constant on the double cosets of $H_n \backslash S_{2n} / H_n$. The permutation $(1,3,\cdots,2n-1)$ that you are interested in belongs to double coset corresponding to the long cycle of $K_{2n}$, and so by a theorem of Diaconis and Landers (see MacDonald's "Symmetric Functions" text),
$$\omega^\lambda_{(n)} = \frac{1}{|H_{n-1}|}\prod_{c \in \lambda}(2a'(c) - l'(c))$$
where $a'(c)$ is the number of cells to the right of $c$, $l'(c)$ is the number of cells below $c$, and the product runs over all cells of $\lambda$ except for the upper-leftmost cell $(1,1)$.
Apologies if this is not what you're interested in.
