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Let $A=\{0,1,\infty,a_1,\ldots,a_n\}$ and $B=\{0,1,\infty,b_1,\ldots,b_n\}$ be subsets of the Riemann sphere.

Let $\sigma$ be an automorphism of the Riemann sphere, i.e., a Möbius transformation, such that $\sigma(A) = B$.

What can one say about $\sigma$?

Example. Suppose that $n=0$. Then $\sigma$ is an automorphism sending $\{0,1,\infty\}$ to $\{0,1,\infty\}$. So $\sigma$ is either the identity map or $z\mapsto \frac{1}{z}$.

Example. Suppose that $n=1$. Under the hypothesis, we have that $b_1 = a_1$. (The cross ratio of $A$ and $B$ should be equal.) So there should be four possibilities for $\sigma$; one of them being the identity map.

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You are missing four further transformations in your first example (you can permute the three points arbitrarily). E.g. $\frac{z}{1-z}$ is another one. – t.b. Jul 21 '11 at 13:50
Closely related. – t.b. Jul 21 '11 at 14:08
The possible transformations sending $\{0,1,\infty\}\to\{0,1,\infty\}$ are $$\left\{z,1-z,\frac{1}{z},\frac{z-1}{z},\frac{1}{1-z},\frac{z}{1-z}\right\}.$$ Under composition these form a group isomorphic to the symmetric group $S_3$. – anon Jul 21 '11 at 14:37
up vote 4 down vote accepted

Three points, $(a,\sigma(a)), (b,\sigma(b)), (c,\sigma(c))\in\mathbb{C}^2$, are enough to wholly determine a Möbius transformation; adding in any more information than this will create an overdetermined situation.

Also, if you have $\{a,b,c\}$ and $\{\sigma(a),\sigma(b),\sigma(c)\}$ without knowledge of which argument goes to what output, you have a total of $3!=6$ different transformations. By making the coefficients of the Möbius transformation unknowns and multiplying through by the denominator, we can solve for the transformation exactly via linear algebra (given a known which-goes-to-what correspondence). Wikipedia has how with an explicit determinant formula. You can then say anything you want about $\sigma$ once you know what precisely it can be.

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