Let $G$ be an infinite simple group. It is requested to prove:
$(i)$ Every $x \in G\setminus \{1\}$ has infinitely many conjugates.
$(ii)$ Every non-trivial proper subgroup $H < G$ has infinitely many conjugates.
which is exercise $3.34$ from J. Rotman's An introduction to the theory of groups.
Here is what has been done:
$(i)$. Let $X = G$. Consider the action $\alpha: G \times X \to X$ given by $\alpha(g, a) = gag^{-1}$, for $a\in X$.
Note that the Orbit of $a \in X$ is $$Orb(a) = \{\alpha(g, a) \in X: g\in G\} = \{gag^{-1} \in X: g\in G\} = a^G$$ which is the conjugacy class of $a$ in $G$, and the stabilizer of $a \in X$ is $$G_a = \{g \in G: \alpha(g, a) = a\} = \{g \in G: ga = ag\} = C_G(a)$$ which is the centralizer of $a$ in $G$.
Then, exists an homomorphism $\phi: G \to S_{\infty}$ such that $\phi(g)(x) = \alpha(g, a)$, for every $g \in G$ and $a \in X$.
By the First Isomorphism Theory, $G/\ker(\phi) \cong Im(\phi)$.
Since $G$ is simple, $\ker(\phi) = \{1\}$ or $\ker(\phi) = G$.
From here it is not clear how to continue. Any valuable idea, hint, or solution is appreciated.
(i) You have $\phi:G\to S_{\mathrm{Orb}(x)}$ with $\phi(g)(a)=gag^{-1}$. If $|\mathrm{Orb}(x)|=n$ were finite, then $S_n$ would be finite, so $\phi$ would have finite image, so its kernel has finite index. But its kernel is normal, so must be trivial or all of $G$. If it were trivial, then $G$ would be isomorphic to a subgroup of $S_n$, hence finite, a contradiction. Thus it is all of $G$, which means $x$ is central, but then $\langle x\rangle$ is a normal (indeed, central) subgroup which must be proper (else $G=\langle x\rangle$ would not be simple, for it would have proper nontrivial normal subgroups $\langle x^n\rangle$ for any $n>1$), a contradiction.
(ii) Same idea, but have $G$ act on subgroups instead of elements by conjugation.