In a comment of this question, t.b. suggests that every compact, homogeneous, totally disconnected, second countable space is either a finite discrete space or the Cantor set; equivalently, it must be a countable power of a finite discrete space. How can this be proven?
Furthermore, $2^\kappa$ is compact, homogeneous, and totally disconnected. Is every compact, homogeneous and totally disconnected space (not necessarily second countable) a power of a finite discrete space?
Best Answer
The answer to the second question is "no", even in the realm of Hausdorff spaces:
Let $X$ be the Double arrow space. It is well-known that $X$ is compact, Hausdorff, zero-dimensional, first countable, but not metrizable. By Theorem 2 in this paper of Alan Dow and Elliot Pearl, $X^\omega$ is homogeneous. Of course, $X^\omega$ is compact, Hausdorff, zero-dimensional (hence totally disconnected) and first countable. Assume $X^\omega$ is homeomorphic to $D^\kappa$ for a discrete, finite space $D$ and a cardinal $\kappa$. By first countability, $\kappa = \omega$, hence $X^\omega$ is metrizable, hence $X$ is metrizable. Contradiction.
Update
I should have thought a little bit more carefully about this before, since there is a much simpler counterexample, which also avoids this very complicated theorem of Dow and Pearl:
The endpoints of the Double arrow space are isolated, hence $Z := X \setminus \{\text{min } X, \text{max } X \}$ is also compact. It is well-known (and not too difficult to see) that $Z$ is homogeneous. Of course, it is also zero-dimensional, Hausdorff, first countable. Hence it is not homeomorphic to $D^\kappa$ for a discrete, finite space $D$.
Anyway, for reference, and since it has already been accepted, I will keep the first example.