I'm trying to prove the following fact:
If $\phi: \mathbb{R}\times \mathbb{S}^1 \to \mathbb{S}^1$, $(t, x)\to \phi^t(x)$ is a family of orientation-preserving circle homeomorphisms, then the orbit $\{\phi^n(x)\}_{n\in\mathbb{N}}$ of $\phi^1$ can't be a Cantor set. In other words, it can never be semiconjugate to an irrational rotation.
If possible I'm not looking for a solution, but rather a hint. What I've got so far is the following: Let's assume $\phi^t(x)\neq x$ for all $x\in\mathbb{S}^1$ and $t\in\mathbb{R}$. We then know that $\{\phi^t(x)\}_{t\in\mathbb{R}} = \mathbb{S}^1$ for every $x\in\mathbb{S}^1$, since the orbit of a group action is connected. Furthermore, it can be assumed that $\{\phi^{p/q}(x)\}_{p/q\in\mathbb{Q}}$ is dense in $\mathbb{S}^1$. However, I don't know how to proceed next, I think it would suffice to show that the orbit of a point $x$ under all integer iterations is a superset (or equal) of the orbit under all rational iterations, but I don't know how to prove that.
Thanks for any help!