A group where every two elements different than 1 are conjugate has order 1 or 2. I need help showing this:
Let G be a finite group such that for every $x$, $y$ in G, $x\neq 1$ and $y\neq 1$, we have that $x$ and $y$ are conjugates. Under those conditions, G must have order 1 or 2.
This is under the topic "actions of groups on sets", but I couldn't figure out a way to start it. Since every element is conjugate, then G must have only one conjugation class, which is itself, but how can this information help?

If $G$ is the trivial group, then the condition holds. So suppose $G$ isn't the trivial group.
$G$ acts on itself by conjugation, i.e. we have an action $G \times G \to G$ where $(g,h) \mapsto ghg^{-1}$. Take an element $x \in G$. Since any two non-identity elements are conjugate, $\mathrm{orb}_G(x) = G \backslash \{e\}$, so $|\mathrm{orb}_G(x)| = |G| - 1$. But the orbit-stabiliser theorem tells us that the size of an orbit divides the order of the group. So $|G|-1 $ divides $|G|$. So $|G|$ must be 2.
