= Solution
Let $G$ be the subgroup preserving the three-point set. Its action gives a <group homomorphism> $\Phi:G\to S_3$. If $\Phi(h)$ is the identity permutation, then $h$ fixes all three points, so part (i) forces $h$ to be the identity Möbius map. Thus the <kernel> is trivial and $\Phi$ is <injective>.
For any permutation $\sigma$, part (i) gives a normalization $g_\sigma$ sending the ordered triple $(z_{\sigma(1)},z_{\sigma(2)},z_{\sigma(3)})$ to $(0,1,\infty)$. If $g$ normalizes the original triple, then $h_\sigma=g_\sigma^{-1}\circ g$ sends $z_j$ to $z_{\sigma(j)}$. Hence every permutation is attained, so $\Phi$ is <surjective>.
$$
\boxed{G\cong S_3.}
$$
This is the setwise stabilizer; its six maps need not fix each point individually.
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