= Solution
Let $T(z)=(az+b)/(cz+d)$, with $ad-bc\ne0$. If $c=0$, then $d\ne0$ and
$$
T(z)=\frac ad z+\frac bd,
$$
a composition of a nonzero scaling and a translation. If $c\ne0$, rearrange the fraction as
$$
T(z)=\frac ac+\frac{bc-ad}{c^2}\frac1{z+d/c}.
$$
Thus every <Möbius transformation> is a composition, in order, of translation by $d/c$, reciprocal inversion, nonzero scaling by $(bc-ad)/c^2$, and translation by $a/c$. All formulas act on the <Riemann sphere>, including zero and infinity. The scaling parameter must be nonzero for that factor to be a <Möbius transformation>.
For three distinct finite points, define the <cross-ratio> normalization
$$
F_z(\zeta)=\frac{(\zeta-z_1)(z_2-z_3)}{(\zeta-z_3)(z_2-z_1)}.
$$
It sends $z_1,z_2,z_3$ to $0,1,\infty$ respectively. If $z_3=\infty$, use $(\zeta-z_1)/(z_2-z_1)$; if $z_1=\infty$, use $(z_2-z_3)/(\zeta-z_3)$; if $z_2=\infty$, use $(\zeta-z_1)/(\zeta-z_3)$. These are the corresponding limits of the same formula and cover all possible triples. Form the analogous normalization $F_w$ of the target triple. Then
$$
\boxed{T=F_w^{-1}\circ F_z}
$$
has all three specified images, proving existence. For uniqueness, the quotient of any two such maps fixes the three source points. Conjugating by $F_z$ gives a <Möbius transformation> fixing $0,1,\infty$. Fixing zero and infinity forces it to be $\lambda\zeta$, and fixing one forces $\lambda=1$. The quotient is identity, so the two maps are equal.
The <Möbius transformations permuting three points> are therefore in one-to-one correspondence with permutations of the triple. For the normalized set the six maps and their ordered images of $(0,1,\infty)$ are
$$
\begin{array}{c|c}
T(z)&(T(0),T(1),T(\infty))\\\hline
z&(0,1,\infty)\\
1-z&(1,0,\infty)\\
1/z&(\infty,1,0)\\
1/(1-z)&(1,\infty,0)\\
z/(z-1)&(0,\infty,1)\\
(z-1)/z&(\infty,0,1).
\end{array}
$$
They realize all six permutations. The action map to the <symmetric group> is a homomorphism, is injective by the three-point uniqueness proof, and is onto by this list. Hence
$$
\boxed{G\cong S_3}.
$$
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