Suppose $f,g,h_1,h_2$ are circle homeomorphisms with $f≠g$ and $fh_i = h_ig$ for $i=1,2$. Does it follow that $h_1 = h_2$?

I restrict $f≠ g$ because I noticed that if

$$f(x) := g(x) := R_\alpha(x) := x+\alpha \mod 1$$ is rotation by $\alpha$ then we could use $h_1 = R_{\beta} = h_2^{-1}$ because rotations commute.

Also, does the situation change if $h_1,h_2$ are just semi-conjugacies?(continuous and surjective)


There are counterexamples when the rotation number of $f$ and $g$ is irrational. If $g$ has an irrational rotation number $\alpha$, then $g$ is semi-conjugate to a rigid rotation $R_\alpha$. Typically the way to construct the topological conjugation $h$ is to pick an orbit of $g$: $\{x_i\}_{i=0}^\infty$ (i.e. $g(x_i) = x_{i+1}$) and for all $n \geq 0$ set $$h(x_n) = k + n\alpha \mod 1$$ Then for any $x_n$ in the orbit we have $R_\alpha \circ h(x_n) = k + (n+1)\alpha \mod 1 = h(x_{n+1}) = h \circ g(x_n)$. If the orbit of $x_0$ is dense then $h$ can be extended to a homeomorphism, and $g$ and $R_\alpha$ are conjugate. If the orbit is not dense, then $h$ can only be extended to a surjection, and $g$ and $R_\alpha$ are semi-conjugate. In either case, we can see our construction works no matter what we choose for $k$ (in other words composition with a rigid rotation produces another topological conjugation). It also did not matter what orbit we chose, which can drastically change our construction of $h$ if the orbit is not dense.

So with $f = R_\alpha$ and $g$ a function which has rotation number $\alpha$, counterexamples exist where $h_1 \neq h_2$. Unfortunately writing down maps with an irrational rotation number which are not simply rigid rotations is somewhat difficult, so I don't have a specific example for you.


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