I set out wondering about the $p$-Sylow subgroups of symmetric group $S_n$. First, if $n=p^k$, the $p$-Sylow is the iterated wreath product $C_p^{\wr k}:=C_p\wr C_p\wr\cdots\wr C_p$. This is the automorphism group of the $k$-level $p$-ary rooted tree (the leaves may be labelled with the numbers $1,\cdots,p^k$). More generally if $n=a_kp^k+\cdots+a_1p+a_0$ is $n$ written in base-$p$ then
$$ P=\prod_{j=0}^k a_j\, C_p^{\wr j} $$
where we interpret $aG=\underbrace{G\times\cdots\times G}_a$ in the above.
We can show $P$ is a $p$-Sylow of $S_n$ by simply calculating its order.
The number of $p$-Sylow subgroups is $S_n/N_{S_n}(P)$, by the orbit-stabilizer theorem, so more generally I was curious what this normalizer is. I think we can say $N_{S_n}(P)=S_{a_0}\times\prod_{j=1}^k a_j N_{S_{p^j}}(C_p^{\wr j})$ using this lemma:
Lemma. If $G\curvearrowright Z$ and $H\le G$ then $Z^H$ is $N_G(H)$-stable. Thus if $G=H\times K$ with $H\le S_X,K\le S_Y$ and $X^H,Y^K$ empty then we can say $N_{S_{X\times Y}}(H\times K)=N_{S_X}(H)\times N_{S_Y}(K)$.
Edit: it seems I have this wrong, should be $(C_p^{\wr j})\wr S_{a_j}$?
It remains to determine $N_{S_{p^k}}(C_p^{\wr k})$. For this, I came to the following question:
Suppose $A\curvearrowright X,B\curvearrowright Y$ and we consider the imprimitive action $A\wr B\curvearrowright X\times Y$. That is, $A^Y\rtimes B$ acts on $X\times Y$ by having $\alpha\in A^Y$ act by $\alpha\cdot(x,y)=(\alpha(y)x,y)$ and $b\in B$ by $b\cdot(x,y)=(x,by)$. Essentially, if we think of $X\times Y$ as an array of ordered pairs with $X$- and $Y$-coordinates, then $A^Y$ has a copy of $A$ for each row to act on the elements within the rows independently, and $B$ permutes the rows themselves.
Question. Under what conditions can we say $N_{S_{X\times Y}}(A\wr B)=N_{S_X}(A)\wr N_{S_Y}(B)$?
