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The answer to your last question is YES. Maybe there is a simpler way to see this, but here is a way:

If $G$ acts freely on $X$ and $\alpha$ is minorising (with $n=2$ if you wish), then there is a zero-dimensional compact space $Z$ of uncountable weight, a suryective map $\pi:Z \to X$ and an action of $G$ on $Z$ commuting with $\pi$ such that $\pi^{-1}\alpha$ is minorising. If $G$ was countable then necessarily $G$ has infinitely many (in fact uncountably many) orbits on the clopen subsets of $Z$. Now you can just consider a suitable countable subalgebra of clopens of $Z$ for which you still have infinitely many orbits and a minorising element (still with $n=2$ if you wish).

Edit: The answer to the first question is also YES. If $G$ acts on $X$ then $G \times \mathbb{Z}_2$ acts in a natural way on $Z$ the disjoint union of two Cantor spaces (which is again a Cantor space), so that the $n$ of a minorising subset $\alpha$ of $X$ gets doubled when $\alpha$ is viewed as a subset of one of the copies in $Z$. Unfortunately $G \times \mathbb{Z}_2$ is not simple. Also, I don´t know if one can get $n=3$ (or any odd number).

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The answer to your last question is YES. Maybe there is a simpler way to see this, but here is a way:

If $G$ acts freely on $X$ and $\alpha$ is minorising (with $n=2$ if you wish), then there is a zero-dimensional compact space $Z$ of uncountable weight, a suryective map $\pi:Z \to X$ and an action of $G$ on $Z$ commuting with $\pi$ such that $\pi^{-1}\alpha$ is minorising. If $G$ was countable then necessarily $G$ has infinitely many (in fact uncountably many) orbits on the clopen subsets of $Z$. Now you can just consider a suitable countable subalgebra of clopens of $Z$ for which you still have infinitely many orbits and a minorising element (still with $n=2$ if you wish).