Let $G$ be a finite group and suppose that $f$ is an automorphism of $G$ with order 3 that fixes only the identity element of $G$. Show that $G$ has a single (i. e. normal) $p$-Sylow subgroup for each prime $p$ dividing $|G|$.
Been working for hours and haven't made much progress. I did note that since $f$ considered as a permutation factors as the product of disjoint 3-cycles, 3 divides $|G|$. So my next step is showing that the 3-Sylow subgroup is normal.
Edit: sorry, since $f$ fixes the identity it actually factors into a bunch of 3-cycles and a 1-cycle. So 3 does not divide $|G|$.