$\exists a\in G-H$ such that $aHa^{-1}=H$ Let $G$ be a $p$-group with proper subgroup $H$. Show that there exists an element
$a\in G -H$ such that $a^{-1} Ha = H$
Can you check my proof?
Since $G$ and $H$ are $p$-groups their centers $Z(G)$, $Z(H)$ are nontrivial. The centers satisfy $Z(H) \leq Z(G)$ and $Z(H) = H\cap Z(G)$.
If $Z(H)< Z(G)$ then $\exists a\in Z(G)$ such that $a\in G-H$ and $ \ \ a^{-1}Ha=H$
If $Z(G) = Z(H) \rightarrow H = Z(G)$ every element of $H$ commutes with every element of $G$ so $\exists a\in G-H \ | \ a^{-1}Ha=a^{-1}aH=eH=H$
I feel like i dindnt proof anything. Is there a better alternative, preferably with group actions?

The question's claim is true in any nilpotent group, and for finite groups it is actually equivalent to being nilpotent, but here are the highlights of a proof without using group nilpotency:
Let $\;G\;$ be a finite $\;p$-group, $\;H\lneq G\;$ . Let us define an action of $\;H\;$ on the set $\;X\;$ of left cosets of $\;H\;$ in $\;G\;$ by left shifting:
$$\forall\,h\in H\;,\;\forall\,g\in G\;\;:\;\;\;h\cdot(gH):=(hg)H$$
Since $\;H\;$ is a $\;p$-group, the orbit-staiblizer theorem tells us that every orbit is a divisor of $\;p\;$. But since $\;|\mathcal Orb(H)|=1\;$, there must be another orbit with only one element (why?), and this means
$$\exists\,g\notin H\;\;s.t.\;\;\forall\,h\in H\;,\;\;h\cdot(gH)=gh\iff hgH=gH\iff g^{-1}hg\in H$$
and we're done.
You can also use induction and go into the quotient group $\;G/Z(G)\;$ and etc.
