A proof of the third Sylow theorem I'm trying to prove the third Sylow theorem, in particular that the number of Sylow $p$-subgroups conjugate to a Sylow $p$-subgroup $P$ (Denoted by $N_p$) is $N_p \equiv 1 \pmod p$. The book exercise is pointing at this method:
Show that the action of $P$ on the set of Sylow $p$-subgroups has $P$ as its only fixed point (easy to show). Then use this and the second Sylow theorem to prove the third.
My attempt:
Using the class formula, $$|P|=|Z| + \sum_{H} [P:\mathrm{Stab}_P(H)]$$
where $Z$ is the set of fixed points of the action (which only has $1$ by the above) and the sum runs over the distinct orbits of the action. But from the second Sylow theorem, every Sylow $p$-subgroup is a conjugate of the other, so in particular the sum only has one term. If we denote the set $P$ is acting on by $S$, then by the orbit-stabilizer theorem, $S\cong P / \mathrm{Stab}_P(H)$. But $|S|=[P:\mathrm{Stab}_P(H)]$ is precisely $N_p$, so I get by the class formula that $|P|=|Z| + N_p$ and taking $\mod p$, $N_p\equiv -1 \pmod p$. What went wrong here?
 A: Based on the comments of @janmarqz, I believe I have resolved the problem, so I will post for completeness (and others to check).
Let $S$ be the set of Sylow $p$-subgroups of the finite group $G$ and let $P\in S$ act on the set $S$ by conjugation. 
First I'll show that $P$ is the only fixed point of the action:
Let $H$ be a Sylow $p$-subgroup that is fixed by the action of $P$. This is equivalent to saying that for every $g\in P$, we have $gHg^{-1}=H$. In particular, $P$ is a subgroup of the normalizer $N_G(H)$ of $H$. Since $H$ is normal in its normalizer, then the subgroup $PH$ is normal in $N_G(H)$ by the second isomorphism theorem, and in particular, $|PH/H|=|P/P\cap H|$. The cardinality of the right hand side is a positive power of $p$, therefore $PH$ is a $p$-group. However, $H$ is a maximal $p$-group of $G$ and hence $PH=H$, which in turn implies that $P$ is a subgroup of $H$. By order considerations, $P=H$, hence $P$ is the only fixed point of the action.
Now, the class equation $$|S|=|S_P|+\sum_H [P:\mathrm{Stab}_P(H)]$$ 
Clearly, $|S|=N_p$, and we know that $P$ is the only fixed point of the action, so $|Z|=1$, and necessarily $[P:\mathrm{Stab}_P(H)]>1$. Since $P$ is a $p$-group, then $p$ divides $[P:\mathrm{Stab}_P(H)]$, so taking the class equation $\mod p$ gives the result.
