Verified answer

Since G is a p-group acting on S, |S'| ≣ |S| (mod p), where S' denotes the subset of S fixed by G. Since p ∤ |S| and |S'| ≣ |S| (mod p), |S'| > 0 (otherwise, |S| ≣ |S'| (mod p) ≣ 0 (mod p), implying that p | |S|). Therefore there is an s' ∈ S'. The element s' satisfies gs' = s' for all g ∈ G. Therefore the orbit Gs' = {s'}.