$|G|=p^n$ where $p$ is prime and $n\geq1$. Show that if $G$ acts on a set $X$, and $Y$ is an orbit of this action, then either $|Y| = 1$ or $p$ divides $|Y|$. Show that $|Z(G)| >1$.
By considering the set of elements of $G$ that commute with a fixed element $x\notin Z(G)$, show that $Z(G)$ cannot have order $p^{n-1}$.
for the first part of this question,I can see |Y|=1 or |Y||p follows from orbit and stabilizer theorem. And consider G acts on itself by conjugation, Z(G) is actually the stabilizer. I got this because intended to show Z(G) is non-trivial by showing orbit can't contain all elements but I didn't find a way to work out. I got stuck here and would be thankful if anyone can help.