= Solution
Each element of $H$ permutes the three-point set, giving a <group homomorphism> $\rho:H\to S_3$. To prove injectivity, let an element fix all three points. Conjugate it by the normalizing <Möbius transformation> from part (i). The conjugate fixes $0,1,\infty$. A <Möbius transformation> $z\mapsto(az+b)/(cz+d)$ fixing $0$ has $b=0$, and fixing infinity has $c=0$. It is then $z\mapsto kz$; fixing $1$ forces $k=1$. Thus the conjugate, and hence the original transformation, is the identity. The <kernel of a group homomorphism> is trivial.
For surjectivity, let $(x_1,x_2,x_3)=(\alpha,\beta,\gamma)$ and choose any permutation $\sigma$. Let $g$ normalize the ordered triple $(x_1,x_2,x_3)$ and let $g_\sigma$ normalize $(x_{\sigma(1)},x_{\sigma(2)},x_{\sigma(3)})$. Then $g_\sigma^{-1}g$ sends $x_j$ to $x_{\sigma(j)}$, lies in $H$, and induces the chosen permutation. Thus $\rho$ is onto and
$$
\boxed{H\cong S_3}.
$$
For the normalized set, the six <Möbius transformations permuting three points> are $z$, $1-z$, $1/z$, $1/(1-z)$, $z/(z-1)$, and $(z-1)/z$. Conjugating by $g$ gives the corresponding six maps for the original set.
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