Let $P$ be a group of order p, on $S\_p$ , How can I prove that the cardinality of normalizer of $P$ it's $p(p-1)$ ?

If I compute that the number of conjugates of the group P, it's $
\frac{{n!}}
{{p\left( {p - 1} \right)}}
$ then I'm done, since equals to the index of the normalizer. But I don't know how.