This is related to my previous question here:
This time I will ask something more precise.
Let $G$ be a group acting by homeomorphisms on the standard Cantor set $X$ (or if you prefer, $G$ is acting on the countable atomless Boolean algebra, which can be realised as the clopen subsets of $X$). Say a non-empty clopen subset $\alpha$ of $X$ is minorising (for $G$) if the following holds:
For every non-empty clopen subset $\beta$ of $X$, there is some $g \in G$ such that $g\beta$ contains $\alpha$.
Now suppose that there is a minorising subset $\alpha$, and that $G\alpha$ (that is the set of $g\alpha$ for $g \in G$) covers $X$. Let $n$ be the smallest size of a subcover of $G\alpha$, that is the smallest number of $G$-translates of $\alpha$ needed to cover $X$. It is clear that $n$ is finite (by compactness) and at least $2$, and that $n$ does not depend on the choice of $\alpha$ (beyond ensuring that $\alpha$ is minorising), so it is an invariant of the action.
The condition $n=2$ is equivalent to the condition that every proper non-empty clopen subset is minorising. (This occurs for instance if $G$ consists of every homeomorphism of $X$.)
Questions: Can $n$ be greater than $2$? What if $G$ is simple, and/or every orbit of $G$ on $X$ is dense?
If $n=2$, can $G$ still have infinitely many orbits on the clopen subsets of $X$?
NB: The existence of a minorising subset means that $G$ destroys most of the extra structure I can think of putting on $X$, such as non-trivial metrics and measures. (Can $G$ act by quasi-isometries?) Effectively $\alpha$ is `as small as possible' up to the action of $G$.