Hint: By Burnside's Lemma (aka, the Lemma Which Is Not Burnside's), the order of G must divide the sum of the orders of the orbits.
If G acts on a finite set S, then S can be written as a disjoint union (See Hungerford's algebra p 93)
, where , , and for all i. We see that 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 is not empty and . An orbit of , denoted , has a single element iff . Can you conclude it from here?