How is the Julia set of $fg$ related to the Julia set of $gf$? Let $f$ and $g$ be complex rational functions (of degree $\geq 2$ if that helps).  What can be said about the relationship between $J(fg)$ and $J(gf)$, the Julia sets of the composite functions $f \circ g$ and $g \circ f$?
If I'm not mistaken, $f$ restricts to a map $J(gf) \to J(fg)$, and $g$ restricts to a map $J(fg) \to J(gf)$.  So $J(fg)$ and $J(gf)$ surject onto each other in a particular way (and, indeed, in a way that commutes with the actions of $fg$ on $J(fg)$ and $gf$ on $J(gf)$).  Since Julia sets are completely invariant, the restricted map $f: J(gf) \to J(fg)$ is $deg(f)$-to-one, and similarly the other way round.  
So there's some kind of relationship between the two sets.
If $f$ or $g$ has degree one then $J(fg)$ and $J(gf)$ are "isomorphic", in the sense that there's a Möbius transformation carrying one onto the other.  Thus, the simplest nontrivial example would be to take $f$ and $g$ to be of degree 2.  I don't know a way of computing, say, the example $f(z) = z^2$ and $g(z) = z^2 + 1$.  That would mean computing the Julia sets of $gf(z) = z^4 + 1$ and $fg(z) = z^4 + 2z^2 + 1$.  
My question isn't completely precise, I'm afraid.  But here are some of the things that I would value in an answer: theorems on what $J(fg)$ and $J(gf)$ have in common, examples showing how different they can be, pictures of $J(fg)$ and $J(gf)$ for particular functions $f$ and $g$, and references to where I can find out more (especially those accessible to a non-specialist).  Thanks. 
 A: I'm not sure if this is helpful, but here is an example.  The following picture shows the filled Julia set for $z^6 - 1$.

and the following picture shows the filled Julia set for $(z^2-1)^3$:

This is the case where $f(z) = z^2 - 1$ and $g(z) = z^3$.  Note that the bottom image is a double cover of the top, while the top image is a triple cover of the bottom.
(These images were produced using Mathematica.)
A: Answer. $J(fg)$ is the full $g$-preimage of $J(gf)$. (And vice versa with interchange of 
$f$ and $g$).
Proof. Let $A=fg$ and $B=gf$. Then we have a semi-conjugacy $gA=Bg$.
Now it is a general fact, that whenever you have such a semi-conjugacy (of rational functions)
the Julia set of $A$ is the $g$-preimage of the Julia set of $B$.
Proof. The semi-conjugacy can be iterated: $gA^n=B^ng$.
Now, $z\in J(A)$, iff the family $gA^n$ is not normal, iff $B^ng$ is not normal, that is
$z\in g^{-1}(J(B))$.
Added on 8.6.12. By the way, this demonstrates an amazing fact: for every $f$ and $g$,
there exist sets $F$ and $G$ such that $G=f^{-1}(F)$ and $F=g^{-1}(G)$.
This looks surprising to me. Finite sets $F$ and $G$ of cardinality greater than $2$
with such properties cannot exist, as a simple count shows.
Are there other examples of such $F$ and $G$ ?
Added on 8.7.12. Let $f$ and $g$ be two rational functions. Let $F$ be a closed set, containing
more than 2 points, and such that $(g^{-1}f^{-1}(F))=F$, then $F$ contains the Julia set of $fg$.
And $J(fg)$ does have this property.
(I am writing compositions $fg=f(g)$.) Trivial, but funny. 
A: Hy
I'm not shure that your restriction map $f: J(gf)\rightarrow J(fg)$ is right.
Insted of looking at Julia set it is equivalent to look at Fatou set where is more natural work with mapings. Since you've  put $f:\{(gf)^n\}_n \rightarrow \{(fg)^n\}_n$ it clearly does not work. Insted you have to use conjugation $f\circ-\circ f^{-1}$, since you have rational function of deg>1 the inverse is not properly defined. So in the neighbour of the periodic points of $gf$ which are not critical (you dont have inverse) for $f$ you get same dynamics. 
As you know fixed points play important role in this theory so you can try to compare fixed points (or periodic) of $fg$ and $gf$.
I think there are some results if you have $g$ an $f$ that commutes and than you compare dimaics of $f$ and $g$ with dynamics of $fg$.
