= Solution
A <Möbius transformation> of the <Riemann sphere> is a map
$$
z\longmapsto\frac{az+b}{cz+d},
\qquad ad-bc\ne0,
$$
with its natural values at the pole and at infinity.
For three distinct points $z_1,z_2,z_3$, define
$$
T_z(\zeta)=
\frac{(\zeta-z_2)(z_1-z_3)}
{(\zeta-z_3)(z_1-z_2)}.
$$
The usual limiting conventions cover an infinite $z_i$. This Möbius transformation sends $(z_1,z_2,z_3)$ to $(1,0,\infty)$. Defining $T_w$ similarly, the map
$$
\boxed{f=T_w^{-1}\circ T_z}
$$
sends $z_i$ to $w_i$. If two Möbius transformations do so, their quotient fixes $0,1,\infty$. A Möbius transformation fixing infinity is affine, and fixing zero and one then makes it the identity. This proves uniqueness.
With this convention, the <cross-ratio> is
$$
[z_1,z_2,z_3,z_4]
=T_z(z_4)
=\frac{(z_4-z_2)(z_1-z_3)}
{(z_4-z_3)(z_1-z_2)}.
$$
Substitution shows that translations, nonzero scalings, and inversion preserve it; since these generate the Möbius group, every Möbius transformation preserves cross-ratios.
Conversely, suppose a <bijection> $f$ of the Riemann sphere preserves every cross-ratio. Let $m$ be the unique Möbius transformation agreeing with $f$ at three chosen points $z_1,z_2,z_3$. For any other $z$, preservation by $f$ and $m$ gives
$$
[z_1,z_2,z_3,z]
=[f(z_1),f(z_2),f(z_3),f(z)]
=[f(z_1),f(z_2),f(z_3),m(z)].
$$
The last coordinate in a cross-ratio with three fixed distinct entries is injective, so $f(z)=m(z)$. The equality already holds at the three base points, hence $f=m$ everywhere and $f$ is Möbius.
Finally, the map $z\mapsto a\overline z+b$ is constant when $a=0$ and therefore is not Möbius. If $a\ne0$, it is bijective and fixes infinity. Were it Möbius, it would have the affine form $\alpha z+\beta$. Equality on real $z$ forces $\alpha=a$ and $\beta=b$, whereas equality at $z=i$ would require $ai=-ai$, contradicting $a\ne0$. Thus
$$
\boxed{\text{there are no such }a,b}.
$$
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