Does $p \mid \frac{|{\rm{Stab}}(Q)|}{|\bigcap_{P\in {\rm{Syl}}_p(G)}{\rm{Stab}}(P)|}, \space\forall Q \in \operatorname{Syl}_p(G)$? Ancillary to the main problem I'm trying to solve, I've come up to a partial result whose generalization would read as follows:

Let $G$ be a finite group, $p$ a prime divisor of $|G|$ and $\operatorname{Syl}_p(G)$ the set of the Sylow $p$-subgroups of $G$. Assume further that $|\operatorname{Syl}_p(G)|>1$. With reference to the transitive action of $G$ by conjugacy on $\operatorname{Syl}_p(G)$, the following holds:
$$p \mid \frac{|{\rm{Stab}}(Q)|}{|\bigcap_{P\in {\rm{Syl}}_p(G)}{\rm{Stab}}(P)|}, \space\forall Q \in \operatorname{Syl}_p(G) \tag 1$$

So far I couldn't prove it nor find a counterexample.

Just for the records, I'm using $(1)$ to prove that, if $G$ has eight $7$-Sylow subgroups, then $G$ has a normal subgroup $N$ such that $56$ divides $[G:N]$.

Let $|G| = p^km$ withe $p$ not dividing $m$. Since $Q \in {\rm Syl}_p(G)$, we have $p^k$ divides $|{\rm Stab}(Q)| = |N_G(Q)|$.
If the result was false then $p^k$ would also divide $|R|$ where $R$ is the intersection of the normalizers of all Sylow $p$-subgroups of $G$. But then $R$ would contain a Sylow $p$-subgroup, say $P$, of $G$, and then you would have $P$ normalizing all Sylow $p$-subgroups, which is impossible because one Sylow $p$-subgroup cannot normalize a different Sylow $p$-subgroup.
