Relationship between order of a group acting on a set and faithfulness

Question

I am trying to prove (or to disprove) that $\forall n,m \in \mathbb{N}$ where $\gcd (n, m) = 1$ and $n, m \geq 2$, $\exists G$ a group such that $|G| = n$ and $\exists X$ a set such that $|X| = m$ such that $G$ acts on $X$ such that there are no trivial orbits (no fixed points). I do not know where to start.

If this is untrue, then is there some restriction such that it becomes true?

Answer 1

I $G$ acts faithfully on a set $X$ then we can define a new action on any set $X\subset Y$ setting $g(a)=a$ for any $a\in Y\setminus X$ and $Y$ can have any cardinality

Answer 2

Let $G=S_3$ and let $X=S_3/\langle (12)\rangle\cup S_3/\langle (123)\rangle$. This set has five elements and the action is faithful.