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Two self-maps $R,S:X\to X$ commute when $R\circ S=S\circ R$. Induction then gives $R^n\circ S=S\circ R^n$ and $R^n\circ S^m=S^m\circ R^n$ for all nonnegative integers $m,n$. Sharing a fixed point or invariant set does not by itself imply commutation. For rational sphere maps of degrees at least two, commutation does imply equality of their Fatou sets and Julia sets, by the commuting rational maps of degree at least two have the same Julia set theorem.

Use the normalized chordal metricUnder stereographic projection it is half the Euclidean chord distance on the unit sphere, which proves the metric axioms and the agreement with the usual sphere topology. If $\rho$ is spherical geodesic distance, then $\rho =2\arcsin \chi$, so $2\chi \le \rho \le \pi \chi$.

A rational map is holomorphic, hence smooth, as a map of compact spheres. At a pole use $1/R$ as target coordinate, and at infinity use $1/z$ as source coordinate; neither is a singularity of the sphere map. Its differential in the round metric has a finite maximum norm $L$. Integrating its norm along a minimizing spherical geodesic gives $\rho (Rz,Rw)\le L\rho (z,w)$. The preceding comparisons implyThis proves that the rational map is Lipschitz in the chordal metric. For a constant map take constant zero; for a nonconstant map the bound is finite by compactness. In finite coordinates the differential norm is $R^{\#}(z)=|R^{\prime }(z)|(1+|z{|}^2)/(1+|R(z){|}^2)$, with its continuous sphere extension.

Now suppose $R,S$ are commuting maps. We first prove $S(F(R))\subseteq F(R)$. The family $\{R^n\}$ is locally equicontinuous on $F(R)$, by the normality equivalence in question 1. Commutation and the Lipschitz estimate give, for all $n$,For $x_0\in F(R)$ choose a source ball within that open set, small enough to make the right side less than any prescribed $ϵ$ uniformly in $n$. Holomorphic openness makes its image contain a ball about $S(x_0)$. Every point $w$ of that target ball has some preimage $x$ in the chosen source ball, so the same inequality proves equicontinuity of the iterates of $R$ at $S(x_0)$. Repeating at each point of the image gives local normality on that image. Thus the stated inclusion holds even where $S$ has a rational critical point.

Carefully state the permitted Montel theorem: a family of meromorphic maps on an open domain, all omitting the same three distinct values of the Riemann sphere, is normal for locally uniform spherical convergence. Since $\mathrm{deg}R\ge 2$, question 1 shows that $J(R)$ contains at least three distinct points. The forward inclusion just proved means every iterate of $S$ maps $F(R)$ into itself and therefore omits those three points there. Montel gives $F(R)\subseteq F(S)$. Exchanging the roles, using $\mathrm{deg}S\ge 2$, gives the opposite inclusion. ConsequentlyThis proves that commuting rational maps of degree at least two have the same Julia set without assuming either is an iterate of the other.

For commuting maps of degree one with different Julia sets, takeTheir two compositions are the identity, and the Julia sets follow from the expanding and contracting Möbius cases in question 1.

For equal Julia sets without commutation, takeBoth affine maps contract toward their finite fixed points and have a repelling periodic point at infinity. But $R\circ S=z/4+1/2$, while $S\circ R=z/4+1$, so they do not commute.

Finally the distinct Chebyshev polynomialshave degrees two and three and neither is a pure power of $z$. Direct expansion gives both compositions equal to $32z^6-48z^4+18z^2-1$. Equivalently $T_m(T_n(z))=T_{mn}(z)$, first seen at $z=\cos \theta$ and then as a polynomial identity. This provides the requested higher-degree pair.