 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.This is the setwise stabilizer; its six maps need not fix each point individually.