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By an analytic isomorphism of $\Omega$, I mean an analytic function $\Omega \to \Omega$, with an analytic inverse.

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Can you extend the function to the whole sphere by continuity? –  Mariano Suárez-Alvarez Oct 22 '12 at 8:47
    
@MarianoSuárez-Alvarez Thanks. You mean I should look at analytic isomorphisms of the sphere (that is Mobius transformations), that preserve the set $\{p_1,\ldots,p_n \}$. –  Teddy Oct 22 '12 at 9:56

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Suppose $f\colon \Omega \to \Omega$ is an analytic isomorphism.

Claim: There exists an analytic isomorphism $\bar{f} \colon \hat{\mathbb{C}} \to \hat{\mathbb{C}}$, such that $\bar{f} \restriction \Omega = f$.

Partial Proof: Essential singularities of $f$ do not exist, for if $p$ is an essential singularity, then $f$ cannot be injective in any $p$-environment (by Picard's big theorem), contradicting the supposed injectivity of $f$.

This means $f$ is meromorphic, and any such function is actually a holomorphic function to $\hat{\mathbb{C}}$. So we may extend $f$ analytically to its poles.

A single pole is possible, but two poles will again contradict $f$ being injective. For suppose $p_1 \ne p_2$ are both poles. Take $U_1 \ni p_1$ and $U_2\ni p_2$ to be open environments. By the open mapping theorem, they are mapped to open environments of $\infty$, which surely intersect.

Assume $p_i$ is not a pole. Then either it is a removable singularity, or it's not a singularity at all. So $f$ may be analytically extended to $p_i$.

Now $\bar f$ is injective by an argument similar to the one using the open mapping theorem. It must be onto since the finite set $\Delta := \{p_1,\ldots, p_n, \infty \}$ must be mapped injectively into itself, which means onto itself.

The analytic isomorphisms of $\hat{\mathbb{C}}$ are exactly the linear fractional (a.k.a. Möbius) transformations (LFTs henceforth). So clearly we may form a bijection between set of analytic isomorphisms $\Omega \to \Omega$ on the one hand, and the set of LFTs which preserve the set $\Delta$ on the other hand.

The latter set is somewhat easier to describe. In case $n=2$ ($\mathbb{C}$ is punctured twice), there exists a unique LFT which maps $\{p_1,p_2,\infty\}$ to a permutation thereof. There are 6 permutations, and therefore 6 such LFTs.

In case $n>2$, one may construct the ${n+1 \choose 3}^2$ LFTs that carry any three distinct points of $\Delta$, to any other three distinct points of the same set, and pick out those that preserve $\Delta$.

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