quasiconformal "automorphism" groups of Julia sets To motivate this question, let me begin with a picture:

Each letter labels a "blob" of this quartic Julia set. (is there a technical term for these parts?). Because of resolution limitations I haven't been able to color and label every blob. 
The only answer to the the MO question Symmetries of Julia sets for $z^2 - c$ mentions that “This means that there is a quasi-conformal map (thus of bounded distortion) which maps parts of the Julia set to the whole.” The transformation that this labelling seems to suggest maps the entire Julia set to itself. 


*

*Does the set of quasiconformal transformations $\Xi$ (as motivated by the one above) which map the Julia set to itself form a group? I am aware that the quasiconformal transformation is not a quasiconformal conjugacy -- while it fixes two points in the plane it also fixes the entire shape of the Julia set)

*If it exists, Is $\Xi$ the discrete subgroup of a continuous group? (I want to make a movie of such a transformation as this)

*What relation, if any, exists between $\Xi$ and Teichmüller space?
The reason that automorphism is in quotes in the question title is because the phrase "quasiconformal automorphism groups" was used in this paper
EDIT: Here are two more of these transformations, applied to the Douady Rabbit: 

The first one simply rotates around a vertex:

In the second, the point that is fixed is inside the mauve Fatou
component. 

 A: Why use a quartic Julia set at all?
Consider the famous "basilica" Julia set; i.e., the Julia set of $z^2-1$. 

Source of image: Prokofiev, Wikimedia commons, 
http://commons.wikimedia.org/wiki/File:Julia_z2-1.png
There is a cycle of two periodic Fatou components: One contains the critical point $0$, the other the critical value $-1$ (which in turn is mapped back to zero). These are connected via a fixed point, which is commonly denoted $\alpha$. (Here $\alpha=\frac{1 - \sqrt{5}}{2}$.) Let $K_1$ be the part of the filled Julia set to the left of $\alpha$ and $K_2$ the part to the right of $\alpha$. Then $f$ maps a neighborhood of $K_1$ conformally to $K_2$. Let us denote this map by $\phi:K_1\to K_2$. Then 
$$ \psi : K \to K; z\mapsto\begin{cases} \phi(z) & \text{if } z\in K_1 \\ \phi^{-1}(z) & \text{if }z\in K_2\end{cases}$$
is a homeomorphism of $K$ (the filled Julia set) to itself. Using the fact that $K_1$ and $K_2$ do not touch tangentially at $\alpha$, it is easy to see that the map extends to a quasiconformal homeomorphism of the plane. Alternatively, consider the corresponding self-map of the lamination: 

Source of image: Adam majewski, Wikimedia Commons, http://upload.wikimedia.org/wikipedia/commons/9/9e/Basilica_lamination.png
The fixed point $\alpha$ corresponds to the "characteristic leaf" connecting angles $1/3$ and $2/3$ (where we think of the circle as $\mathbb{R}/\mathbb{Z}$. The map in question is given by
$$ \Psi(x) = \begin{cases}2x-k & \text{if } x\in [k+1/3,k+2/3], k\in\mathbb{Z} \\
                          (x+k+1)/2 & \text{if } x\in (k-1/3,k+1/3), k\in\mathbb{Z}. \end{cases}.$$
This map is clearly a quasisymmetric self-map of the circle, and hence extends to a quasiconformal self-map of the complement of the unit disk. This gives rise to a quasiconformal self-map of the complement of $K$ that agrees with $\psi$ on the boundary of $K$. By the classical glueing lemma for quasiconformal mappings, the combined map is quasiconformal near every point apart possibly from $\alpha$, and hence quasiconformal everywhere.
Presumably the same is true for the example you post, although since you do not say exactly which map you are considering, it is impossible to tell.
To answer the three questions posed by you:


*

*Of course the set of quasiconformal maps that take the Julia set to itself forms a group. This is trivial because the composition of two such maps again preserves the Julia set, and of course the same is true for inverses. However, note that the quasiconformal dilatation can increase under composition.

*Of course this is a subgroup of the group of quasiconformal self-maps of the plane. The subgroup is not discrete because you can continuously deform the quasiconformal map on the Fatou components. If we identify two maps when they take the same values on the Julia set, the question because more sensible. However, even then I believe the answer is negative. Indeed, it should be possible to define a quasiconformal isomorphism that fixes $K_2$ and "rotates" the bulbs around on the other side. By doing this kind of transformation on one of the iterated preimages of $K_2$ that is very far to the left (or right) of the Julia set, and keeping the rest of the set fixed, we obtain a map that is very close to the identity on the Julia set. (Of course details need to be checked.)

*I do not understand your third question. What Riemann surface are you considering?
EDIT. Lyubich and Merenkov have recently given a complete description of the group of orientation-preserving quasisymmetries of the basilica Julia set. In contrast, for certain rational maps, they previously showed together with Bonk that every quasisymmetry is a Möbius transformation.
A: These blobs are called connected components of the Fatou set, which is the complement to the julia set (the julia set is the boundary of your blobs, as you probably know already). It is known that for rational maps (iterating maps of the form $f(x)=p(x)/q(x)$ where p,q are polynomials) result in 1, 2 or infinitely many components. 
Since the julia set is both forward and backward invariant under f, 
the components are as well. Hence, the function f, and its inverse, are such functions (you'll probably need to choose a branch of the inverse of f for each Fatou component). Related to this, look at this wiki page about a famous theorem related to this: http://en.wikipedia.org/wiki/No_wandering_domain_theorem
This might also be of interest: http://en.wikipedia.org/wiki/Classification_of_Fatou_components
I am unsure if there are any other reasonable maps that maps fatou components to fatou components.
