A transitive action on a specific set Let $G$ be a finite group and $\lambda\in G$, consider a set $$D^{p+*}_{G}(\lambda):=\{P\in S^{p+*}_G|P^{\lambda}=P=[P,\lambda]\}$$ where:
$S^{p+*}_{G}$ denotes the set consisting of all non-trivial $p$-subgroups of $G$ with conjugation actions by $G$, and
$[P,\lambda]$ denotes the subgroup of $G$ generated by commutators $[x,\lambda]$ where $x\in P$.
It is clear that the centralizer $C_{G}(\lambda)$ acts on the set $D^{p+*}_{G}(\lambda)$.
Suppose that we take $p=2$, $G=\Sigma_n$, $n\geq 6$, and $\lambda=(1,2,3)$ a $3$-cycle.
Then it seems like in this case the action $C_{\Sigma_n}((1,2,3))$ on $D^{2+*}_{\Sigma_n}((1,2,3))$ is transitive with the stabilizer subgroup isomorphic to $C_{\Sigma_{n-1}}((1,2,3))$.
Does anyone know how to prove this statement?
 A: I will denote $D_{\Sigma_n}^{2+*}((1,2,3))$ by $D$.
We claim that $D$ consists of $n-3$ Klein $4$-groups:
$$D = \{ \langle (1,2)(3,k),(1,3)(2,k) \rangle : 4 \le k \le n \}$$
and the result follows easily from this.
To see this, let $P \in D$. Then the condition $[P,\lambda]=P$ implies that $P$ cannot fix the three points 1,2,3, and then $\lambda \in N_G(P)$ forces $1,2$ and $3$ to be in the same orbit $\Delta$ of $P$.
Then $P$ cannot have any other nontrivial orbits, since otherwise $\lambda$ would centralize the action of $P$ on that orbit, which is a quotient group of $P$, and then we would have $[P,\lambda] \le N$, where $N$ is the kernel of the action of $P$ on that orbit,  contradicting $[P,\lambda]=P$.
Now consider the nontrivial orbits of the centre $Z(P)$ of $P$. These are subsets of $\Delta$ all of the same size, and since $\lambda \in N_G(Z(P))$, we find that $1,2,3$ are in the same orbit $\Delta'$ of $Z(P)$.
If $Z(P)$ had any other nontrivial orbits, then we could consider the induced actions of $P$ and $\lambda$ on these orbits. Since $\lambda$ would act as the identity in this induced action, we could not have $[P,\lambda]=P$, so $\Delta'=\Delta$, and $Z(P)$ is transitive on $\Delta$.
But the action of $Z(P)$ on $\Delta$ must be regular, and it is self-centralizing in ${\rm Sym}(\Delta)$, so $P=Z(P)$ and $P$ is abelian.
Then by a standard result, we have $P = [P,\lambda] \times C_P(\lambda)$, so $C_P(\lambda)=1$. But $|C_P(\Delta)| = |\Delta|-3$, so $|\Delta|=4$, and the claim follows easily.
