Second Sylow theorem states that all Sylow $p$-subgroups are conjugate. But reviewing my proof it seems to me that we also prove that all the conjugates of a Sylow $p$-subgroup are Sylow $p$-subgroups.
I can include the proof if needed but can anybody confirm me this idea? Or give a counterexample?
On
Yes, but that is more or less trivial, once you know that conjugation by a fixed element is an automorphism.
If $H$ and $H'$ are conjugate subgroups of $G$, then the inner automorphism of $G$ that takes $H$ to $H'$ will also take any subgroup between $H$ and $G$ to an isomorphic subgroup between $H'$ and $G$. Therefore if $H'$ is a maximal $p$-subgroup, and $H$ will also be a maximal $p$-subgroup, and vice versa.
In general if $H$ is a subgroup of $G$ and $g\in G$, then we have $|gHg^{-1}|=|H|$, so it follows that the conjugates of a Sylow subgroup are also Sylow subgroups.