Is there an entire function with domains for which $f(A)=B$ and $f(B)=A$?

Question

Let $f$ be an entire function. Suppose that there exist two nonempty disjoint, open, connected non-empty sets $A,B$ in the plane such that $f(A)=B$ and $f(B)=A$.

Does it follow that $f$ is linear?

Equivalently, if a meromorphic function satisfies this condition is it necessarily an automorphism?

Neither of the conditions of disjointness and openness can be dropped, of course. I tried to see if results in dynamics about 2-periodic domains apply, but they usually only regard Fatou components or are otherwise not suitable. But it does seem like a question simple enough that it "ought to" be amenable to such machinery.

Any ideas?

Answer 1

The conclusion does not hold, not even for polynomials. If $z_0$ is an attracting fixed point of $f \circ f$ (but not a fixed point of $f$) and $A$ the component of the Fatou set containing $z_0$, then $B = f(A)$ is disjoint from $A$ with $f(B) = A$.

A concrete example is $f(z) = z^2 - 1$ with $f(0) = -1$, $f(-1) = 0$, and $A, B$ the components of the Fatou set containing $0$ and $-1$, respectively.

Here is an image of the Julia set of $z^2-1$ (Attribution: Prokofiev / Public domain):

The Fatou component in the center contains $z=0$ and the next one on the left contains $z=-1$.

Answer 2

Not a full answer but too much for a comment.

If $A$ and $B$ have elements on thin hairlike parts of the their full set then we can conclude this :

If $f$ is entire then $f$ is completely determined by its taylor expansion. So you want $f(A) = B, f(B) = A$.

This implies

$$f(f(A)) = A$$ $$f(f(B)) = B$$

Since $A,B$ are connected they contain uncountable elements.

Therefore when you expand your taylor around elements of $A$ or around elements of $B$ that are thin you get an uncountable number of fixpoints for $f$ iterated twice.

This implies that $f(f(z)) = z$ for thin elements of $A$ or $B$.

Since $f$ is entire this implies that $f(z) = z$.

We can conclude that this implies that $A,B$ can not have thin hairlike parts or else $f(z)=z$.

In fact this extends to locally analytic functions $f(z)$ hence I comment this.