Let $(X, d)$ be a compact metric space and $G$ a discrete group acting on $X$ such that, for each $g\in G$, the mapping $x\mapsto g\cdot x$ defines a homeomorphism on $X$
Is it true that the existence of an infinite $G$-invariant set will imply that the group is cyclic?
Given $G$ is cyclic, for instance $G = \mathbb{Z}$, the action of $\mathbb{Z}$ acting on the unit circle does not have any invariant sets. Here is my attempt on the other direction: Assume $S$ is a $G$-invariant set and we fix $g\in G$. Then $\langle\,g\,\rangle\cdot S = S$ because for each $s\in S$, we have $g\cdot s\in S$, and hence $g^{-1}\cdot(g\cdot s) = s$. Because $g$ is arbitrarily fixed and $G\cdot x$ is dense in $X$ for each $x\in X$, for a fixed $s\in S$, we have:
$$ \forall\,h\in G,\hspace{0.2cm}\langle\,h\,\rangle\cdot s\subseteq S = \langle\,g\,\rangle\cdot s\,\implies\, \langle\,g\,\rangle\cdot s = G\cdot s $$
Then, if there exists $h\notin\langle\,g\,\rangle$, what conditions do I need to conclude that $h$ is some $\epsilon$ away from infinitely many elements in $\langle\,g\,\rangle\cdot s$, or can I always conclude that?
The first question is inspired by the following particular case: when $G = \mathcal{F}_n$, the free group with $n$ generators and $X$ is the Gromov boundary of $\mathcal{F}_n$ (we can view $X$ as the set of words with infinite length), given $x, y\in X$, define a metric $\delta(x, y) = \operatorname{exp}(-\vert\,x\wedge y\,\vert)$ where $\vert\,x\wedge y\,\vert$ is the first $k$-th digit where $x$ differs from $y$. Also, for each $g\in\mathcal{F}_n$ and $x\in X$, define $g\cdot x = gx$. In this case $(X, \delta)$ will be a compact metric space and one can show that the existence of a finite invariant set in $X$ implies that $G$ will be cyclic.
Therefore, I wonder if this will be true for an arbitrary discrete group acting on a compact metric space. Any hints to either question will be appreciated.
Update: thanks for YCor's inputs from MO, in the first question, "infinite" cannot be replaced by "finite". A counter-example will be that the group of homeomorphisms of the unit circle $G$ acting on the unit circle $\mathbb{T}$. In this case, for each $S\subseteq\mathbb{T}$ and any subgroup $H\leq G$, $H\cdot S\subseteq S$ iff $H\cdot S=S$. Hence any subgroups of $G$ that fixes a finite set must be uncountable and hence such a group will not be cyclic.
As pointed out by Moishe Kohan, the proposition is apparently false. It is not true that "..., the existence of an infinite $G$-invariant set will imply that the group is cyclic."
A trivial counterexample
Let $X=[-1,0]\cup\{1,2,3\}$ with the inherited metric of $\Bbb R$. $X$ is compact.
Let $G$ be the set of all $6$ auto-homeomorphism of $X$ that fix $[-1,0]$ but permute $1,2,3$.
While all infinitely many numbers in $[-1,0]$ are fixed by $G$, $G$ is not cyclic since $G$, which is the symmetric group over three elements, is not even commutative.
This counterexample is trivial in the sense that $X$ is not connected.
A non-trivial counterexample
Here is a counterexample where $X$ is connected.
Let $X$ be the graph above, a union of $4$ line segments in a $\Bbb R^2$ with the inherited metric. $X$ is compact. $\overline {OA},\overline {OB},\overline {OC}$ can be viewed as three identical copies of some line segment $S$.
Let $G$ be the set of all $6$ auto-homeomorphism of $X$ that fix $\overline{OI}$ but permute the three line segments (that induce the identical map on $S$).
While all infinitely many points on $\overline {OI}$ are fixed by $G$, $G$ is not cyclic since $G$ is not commutive as before.