Proof of non-existence of topological semi-conjugacy between $R_{\alpha}$ and $R_{2 \alpha}$

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I am just starting to study dynamical systems so this might not be a particularly complicated question, but I've been scratching my head for days without much success.

I am looking at the map of rotations of the circle $R_\alpha:S^1\rightarrow S^1$, $R_\alpha(x)=x+\alpha$ mod 1, with irrational $\alpha$. My aim is to show that $R_{2\alpha}$ is a factor of $R_\alpha$ (as in, there is a topological semi-conjugacy $\psi$ with $\psi\circ R_\alpha=R_{2\alpha}\circ\psi$), but that the converse is not true (as in, there is no topological semi-conjugacy $\psi$ with $\psi\circ R_{2\alpha}=R_{\alpha}\circ\psi$). As a hint, I'm told to use the fact that all orbits under an irrational rotation are dense.

The first part was easy, just taking $\psi$ to be the doubling map $\psi(x)=2x$ mod 1 works fine. I'm struggling with the second part, especially because the hint confuses me. Every point in the circle has a dense orbit under an irrational rotation, but both $R_\alpha$ and $R_{2 \alpha}$ are irrational rotations, so I can't see how that would help here. I've tried starting from the definition of dense orbit and somehow using the semi-conjugacy property to reach some sort of contradiciton, but so far I've found nothing meaningful. I'd really appreciate any hints to get me started on this.

Thanks!

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Rotations of the circle are uniquely ergodic, minimal and preserve the usual metric.

Suppose there is a conjugacy $\Phi$ between two rotations $R_\alpha$ and $R_\beta$. By potentially composing $\Phi$ with a rotation we can w.l.o.g. assume that $\Phi(1)=1$. Let us count how often we cross the point $1$. For that we consider the hitting times of the intervals $(\alpha,1]$ and $(\beta,1]$ respectively. By the Birkhoff ergodic theorem the density of those hitting times is $\alpha$ and $\beta$ respectively. (Look up the Keyword "Sturmian Sequences" for more information)

Proof that a conjugacy must preserve those times of crossing $1$.

Conclude that $\alpha = \beta$.