Let $G$ be a finite group, $p$ a prime dividing the order of $G$, and let $S=\{ P: P \text{ is a Sylow }p \text{-subgroup of }G\}$. Let $x\in G.$ Then $S=S^{x}.$
Proof. Since the elements of $S^{x}$ are Sylow subgroups of $G$, then $S^{x} \subseteq S$ .Define the function $f:S \rightarrow S^{x}$ by $f(P)=P^{x}$. Let $P_{1}, P_{2}$ be elements of $S$.
$f$ is well defined. If $P_{1}=P_{2}$, then $P_{1}^{x}=P_{2}^{x}$ and so $f(P_{1})=f(P_{2})$.
$f$ is one-to-one. If $f(P_{1})=f(P_{2})$, then $P_{1}^{x}=P_{2}^{x}$ and so $P_{1}=P_{2}$.
$f$ is onto. Let $P^{x} \in S^{x}$. Then we have $P \in S$ such that $f(P)=P^{x}$.
Thus $S=S^{x}$
Is the above true?