= Solution
For a <holomorphic map>, <conformality> means preservation of oriented angles locally; it holds at every point of $G$ exactly when $f'(z)\ne0$ there. A <Möbius transformation> is $M(z)=(az+b)/(cz+d)$ with $ad-bc\ne0$, extended to the <Riemann sphere> at its pole and at infinity.
For three distinct finite points define
$$
T_z(z)=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)},
$$
which sends $z_1,z_2,z_3$ to $0,1,\infty$. If one point is infinite, take the corresponding limit: for $z_3=\infty$ use $(z-z_1)/(z_2-z_1)$; for $z_1=\infty$ use $(z_2-z_3)/(z-z_3)$; for $z_2=\infty$ use $(z-z_1)/(z-z_3)$. Construct $T_w$ similarly. Then \b[$M=T_w^{-1}\circ T_z$] has the required three images.
For uniqueness, normalize any two proposed maps by these same transformations. Composing one normalized map with the inverse of the other fixes $0,1,\infty$. A Möbius map fixing infinity has $c=0$, one fixing zero then has $b=0$, and fixing one forces $a=d$ up to the irrelevant common nonzero scalar. It is the identity. This proves <sharp three-transitivity of Möbius transformations>.
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