Equicontinuity and orbits of compact open sets Let $X$ be a compact zero-dimensional space, let $S \subseteq \mathrm{Homeo}(X)$ and let $U$ be a compact open subset of $X$.  Suppose that $s^{-1} \in S$ for all $s \in S$, and that $S$ restricts to an equicontinuous family of functions from $U$ to $X$.  Is the set $\{sU \mid s \in S\}$ finite?
My thinking is that $S$ should be relatively compact in some sense, as far as $U$ is concerned (by an Arzelà–Ascoli type theorem) and the setwise stabilizer of $U$ is an open subgroup of $\mathrm{Homeo}(X)$ in the compact-open topology, so $\{tU \mid t \in T\}$ is finite for any compact subset $T$ of $\mathrm{Homeo}(X)$.
 A: I couldn't quite answer the question as posed, but thanks to John Griesmer's hint I managed to get a positive answer with slightly different (and perhaps more 'natural') hypotheses, so I will put something here so that the question doesn't remain unanswered.
Claim: Let $X$ be a compact zero-dimensional space, let $S \subseteq \mathrm{Homeo}(X)$ and let $U$ be a clopen subset of $X$.  Suppose that $S^{-1}:=\{s^{-1} \mid s \in S\}$ is equicontinuous on $Y = \overline{SU}$.  Then $\{sU \mid s \in S\}$ is finite.
Proof: We regard $S^{-1}$ as a family of functions from $Y$ to $X$ by restricting the domain.  This family of functions is equicontinuous and uniformly bounded, so by the Arzelà–Ascoli theorem (see e.g. Willard, General Topology, Theorem 43.15), $S^{-1}$ has compact closure in $C(Y,X)$.  We then have a continuous map $\theta$ from $C(Y,X)$ to $C(Y,\mathbb{R})$ given by $\theta(f)= 1_{U} \circ f$, so $\theta(S^{-1})$ has compact closure in $C(Y,\mathbb{R})$.  Since elements of $\theta(S^{-1})$ only take values in $\{0,1\}$, $\theta(S^{-1})$ is discrete in $C(Y,\mathbb{R})$, hence finite.  Finally $\theta(S^{-1})$ coincides with the subset $\{1_{sU} \mid s \in S\}$ of $C(Y,\mathbb{R})$, so $\{sU \mid s \in S\}$ is finite.
