Let $G$ act on set $S$ with $|G|=3^n, |S|=2021$. Prove $\exists s_1,s_2, {\rm Stab}_G(s_1)={\rm Stab}_G(s_2) = G$ not in the same orbit

Question

I have a group $G$ acting on set $S$. $$f : G \times S \rightarrow S$$ $|G| = 3^n$ and $|S| = 2021$. I need to show that there exists two elements $s_1$ and $s_2$, such that ${\rm Stab}_G(s_1) ={\rm Stab}_G(s_2) = G$ and $s_1$ and $s_2$ are not on the same orbit!

I was trying to get and answer using the Dirichlet principle. It didn't work out.

Then I was trying to find some connection with the symmetric group $S_{2021}$. But I can't seem to solve it.

Any tips?

Answer 1

By the orbit-stabilizer theorem, each orbit's size must be either a proper power of $3$, or $1$. Moreover, $S$ is partitioned into the set of orbits. Therefore, since $2021\equiv 2 \pmod 3$, there must be at least two singleton orbits (whose stabilizers are then the whole $G$).