Let $P$ be a normal sylow $p$-subgroup of a finite group $G$.
Since $P$ is normal it is the unique sylow $p$-subgroup.
I would like to say if $\phi$ is an automorphism then $\phi(P)$ is also a sylow $p$-subgroup. Then uniqueness would finish the proof. But is that true?
Does an automorphism of a group always send subgroups to subgroups of the same order?
Let $\phi:G \to G$ be the automorphism. We'll prove $\phi(P)$ is a $p$-Sylow Subgroup of $G$.
$\phi$ takes identity to itself:
For any $e_G \in G$, the identity in G, $\phi(e_G.e_G)=\phi(e_G)\phi(e_G)$ which proves the result.(from cancellation law in a group)
$\phi$ takes inverses to inverses:
Use the fact that $gg^{-1}=e_G$ for any $g \in G$. Do a computation similar to the one above.
$\phi$ takes subgroups to subgroups:
Let $H \leq G$. We intend to prove $\phi(H)$ is a subgroup. Use the 'lemma' we have proved before and verify the subgroup criterion (that $\phi(H)$ is closed under multiplication and inverses. )
Now, by one of my comments above, (in fact by just using the bijectivity of the map $\phi$, and by looking at its restriction to $H$), we'll prove that $|\phi(H)|=|H|$.
Note that the definition for two sets to be of same cardinality is that there exists a bijection between them.
So, You are through.
Take any automorphism $f$, then $\#f(P)=\#P$ (because in particular $f$ is a bijection), and since also the image of subgroups (in this case $P$) is a subgroup, you conclude $f(P)$ is a p-sylow, and since you remarked:
Since $P$ is normal it is the unique Sylow $p$ subgroup.
That means $f(P)=P$,
