prove : let $G$ be a finite group and any two maximal subgroups are conjugate then $G$ is cyclic .also can we say $|G|=p^n$ for prime number $p$ ?
If $G$ has only one conjugacy class of maximal subgroup, $M$, and is not itself a $p$-group, then for every prime $p$, $M$ contains a Sylow $p$-subgroup $P$ of $G$. But then by Lagrange, $|M|$ has order divisible by every prime power dividing $|G|$, so $|M|$ is a multiple of $|G|$. Since $|M| \leq |G|$, one must have $|M| = |G|$ and $M = G$, a contradiction.