Let $G$ be a $p$-group. If every two maximal subgroups of $G$ are conjugated, then $G$ is cyclic I already know:
Let $|G|=p^{n}$. Every maximal group is normal, since two maximal subgroups are conjugated, then there is an unique $P$ normal subgroup of $G$ maximal with order $p^{n-1}$.
Also, $G/P\simeq \mathbb{Z}_p$, so there is $x\in G\setminus P$ such that $G/P=\langle x+P\rangle$ and the order of x is p. therefore  $G=P\ltimes \langle x\rangle$
IF $P$ is cyclic, let $y$ be a generator so $e=x^{i}y^{j}$ for some i and j, so $x^{-i}=y^j$. Thus every element it could be generated by $y$, becuse $\langle x^{-i} \rangle=\langle x \rangle \simeq \mathbb{Z}_p$
But IF $P$ is not cyclic? Can I do a kind of inducction? any sugestion?
Thanks!!

I think there is a mistake in the beginning of your argument: you claim that there is $x \in G \setminus P$ such that $G/P$ is generated by $xP$ and the order of $x$ is $p$. This is not true, for example if $G$ is cyclic. (Explicit example: $G$ cyclic of order $4$.)
For the problem you are asking about: you seem to have noticed already that if every maximal subgroup is conjugate in a finite $p$-group $G$, then $G$ has a unique maximal subgroup $P$. Well, now take some element $x \in G \setminus P$. Can you see why $\langle x \rangle = G$?
