Let $G$ be a finite group of odd order $n$. Suppose that $G$ acts on a set $X$ and let $Y\subseteq X$ be an orbit. Show that if $|Y| < n$ then $|Y| \le \frac n3$.
I know that if $G$ is a finite group acting on a set $X$, then every orbit is a finite set and its cardinality divides the order $|G|$ of the group, is this what I'm meant to use to solve this question?
I'm just a little unsure and I cant find anything similar in my notes
HINT: As you suggested, by Orbit Stabilizer Theorem, we have
$$|G| = |O_a|\cdot|G_a|,\ \text{for any }a \in X$$
Thus, for an orbit $Y \subseteq X$, we have $|Y|\ \big{|}\ n=|G|$. If $n$ is odd and $|Y| \ne n$, what can we say about $|Y|$? What is the largest integer of the form $\frac{n}{k}$ if $k \ge 2$ with $k \in \mathbb{Z}^+$ when $n$ is odd?