Sylow implies automorph-conjugate

Statement
In a finite group, any Sylow subgroup is an automorph-conjugate subgroup.

Hands-on proof
Given: A finite group $$G$$, a $$p$$-Sylow subgroup $$S$$

To prove: For any automorphism $$\sigma$$ of $$G$$, $$S$$ and $$\sigma(S)$$ are conjugate

Proof: The key thing to observe is that $$\sigma(S)$$ is also a $$p$$-Sylow subgroup. Hence, $$S$$ and $$\sigma(S)$$ are $$p$$-Sylow subgroups, so by the conjugacy part of Sylow's theorem, they are conjugate.