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

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?

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, $$h_1\circ h_2^{-1}=R_{p-q}.$$