Is it true that if $h\circ R_{\tau}=R_{\tau}\circ h$, then $h$ is a rotation?

Question

Note: Every map mentioned in this problem is a homeomorphism $S^1\to S^1$.

I'm working on a problem from Katok/Hasselblatt, which is

Show that if $f$ is topologically conjugate to an irrational rotation $R_{\tau}$, then the conjugating homeomorphism is unique up to a rotation. That is, if $h_i\circ f=R_{\tau}\circ h_i$ for $i=1,2$, then $h_1\circ h_2^{-1}$ is a rotation.

Now, in the back of the book there is the following hint:

Hint: Show that $h_1\circ h_2^{-1}\circ R_{\tau}=R_{\tau}\circ h_1\circ h_2^{-1}$.

Now, this isn't difficult, because we can write by assumption

$$h_1^{-1}\circ R_{\tau}\circ h_1=f=h_2^{-1}\circ R_{\tau}\circ h_2,$$

from which rearranging gives us $h_1\circ h_2^{-1}\circ R_{\tau}=R_{\tau}\circ h_1\circ h_2^{-1}$.

Now, I have no idea how to see that $h_1\circ h_2^{-1}$ is a rotation matrix from this fact. There doesn't seem to be anything special about $h_1\circ h_2^{-1}$ here, so I guess the problem boils down to

If $h\circ R_{\tau}=R_{\tau}\circ h$, then $h$ is a rotation.

However, I don't know how to show this. I haven't used at all the fact that $\tau$ is irrational, so that must be important. But I don't know how to use that fact. Can somebody help?

Answer 1

Hint: a conjugacy takes orbits to orbits.

And in this case all orbits are dense. So, if you start with one and then with another one, the images, say $p$ and $q$, of the two initial points of the two orbits are sufficient to determine the two conjugacies (you take the image of an orbit and the rest is obtained using densedness).

Now, $p-q$ (seen in the line) is how much you rotate from one orbit to the next, that is, $$h_1\circ h_2^{-1}=R_{p-q}.$$