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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 $\mathrm{Homeo}(X)$.

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    $\begingroup$ You're about one sentence away from a proof. After applying Arzelà-Ascoli, observe that a collection of indicator functions $\{1_{V}:V\in \mathcal V\}$ is compact in the topology induced by the uniform norm on $C(X)$ if and only if $\mathcal V$ is finite. Reason: a sequence of indicator functions converges in the uniform norm if and only if the sequence is eventually constant. $\endgroup$
    – John Griesmer
    Commented Jul 7, 2017 at 16:20
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    $\begingroup$ I see that $\{1_{sU} : s \in S\}$ is compact if and only if it is finite, and it's contained in a continuous image of the closure of $S$ in $C(X,X)$. But I'm not sure exactly where the compactness kicks in, as $S$ is only assumed equicontinuous on $U$. $\endgroup$
    – Colin Reid
    Commented Jul 8, 2017 at 8:58
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    $\begingroup$ I now see how the argument works if we assume $S$ is equicontinuous on $\overline{SU}$. It's not quite what I originally asked, but it is sufficient for the application I had in mind. $\endgroup$
    – Colin Reid
    Commented Jul 9, 2017 at 5:00

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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.

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