Aluffi IV.2.10 suggests the following exercise. Let $P$ be a $p$-Sylow subgroup of a finite group $G$, and act with $P$ by conjugation on the set of $p$-Sylow subgroups of $G$. Show that $P$ is the unique fixed point of this action.
So how do I prove this?
There's also a hint to use a previous exercise I just worked out, claiming that if $H$ is a $p$-subgroup and $H \subset N_G(P)$, then $H \subset P$, but I haven't got to the point where I could see it being useful.
Apparently I had to go get a cup of tea, this got done immediately after that.
So let $P'$ be a $p$-Sylow group such that it is a fixed point of this action. That is, $\forall g \in P : gP'g^{-1} = P'$. But this means precisely that $P \subset N_G(P')$, hence $P \subset P'$, hence $P = P'$.
This now seems so blatantly obvious that I probably should've just deleted my question.