Context. This question is a follow-up of the following one.
Let $n\geq 1$ be an integer, let $d\mid n$, and let $H=\langle\sigma, \tau \rangle\subset S_n,$ where $ \sigma=(1 \ 2 \cdots n), \tau=( 1 \ d+1)$.
According to an exercise of Serre's finite groups, after correction of some typo, one should have $H\simeq (S_{n/d})^d\rtimes \mathbb{Z}/d\mathbb{Z}$, which is confirmed by a CAS for various values of $n$ and $d$.
For $1\leq k\leq d$, let $E_k=\{ k+m d, m=0,\ldots,\dfrac{n}{d}-1\}$.
One may show that $\sigma(E_k)=E_{k+1}$ and $\tau(E_k)=E_k$ (where the indices are taken modulo $n$). In fact,
Then $H$ acts on $X=\{E_1,\ldots,E_d\}$, and induces a morphism $\rho:H\to S_d$. The image of $\rho$ is cyclic of order $d$ (generated by $(1 \ 2 \cdots d)$ by the previous considerations, and a permutation acts like the identity if and only if it lies in $S(E_1)\odot\cdots\odot S(E_d)$.
One may show easily that all the transpositions $(i \ i+\ell d)$ lie in $H$, so the kernel of $\rho$ is exactly $S(E_1)\odot\cdots\odot S(E_d)\simeq (S_{n/d})^d.$
One remains to find is a suitable element $\nu$ of order $d$ in $H$.
Experimentation with a CAS seems to suggest that $\nu=(1 \ 2 \ \cdots d)(d+1 \ d+2 \ \cdots 2d)\cdots ((\frac{n}{d}-1)d+1 \ \cdots n)$ lies in $H$.
I think that this $\nu$ does the job, that is, $H=S(E_1)\odot\cdots\odot S(E_d)\rtimes \langle\nu\rangle$, but I am not able to prove that it lies in $H$ if full generality.
So my question is:
Question. How to prove that $\nu=(1 \ 2 \ \cdots d)(d+1 \ d+2 \ \cdots 2d)\cdots ((\frac{n}{d}-1)d+1 \ \cdots n)$ lies in $H$ ?
I will also be happy to have another solution to prove the isomorphism $H\simeq (S_{n/d})^d\rtimes \mathbb{Z}/d\mathbb{Z}.$