$S=S_0 \cup \bar{x_1} \cup \bar{x_2} \cup \cdots \cup \bar{x_n}$, where $S_0=\{x \in S: g\cdot x=x \text{ for all } g \in G\}$, $|\bar{x_i}|=|G:G_{x_i}|$, and $|\bar{x_i}|>1$ for all i. We see that $|\bar{x_i}|$ is divisible by p by Lagrange's theorem, since a p-group G acts on S. By assumption, S is not divisible by p, so $S_0$ is not empty and $p \nmid|S_0|$. An orbit of $x \in S$, denoted $\bar{x}$, has a single element iff $x \in S_0$. Can you conclude it from here?