= Solution
A nonidentity <Möbius transformation> has either two distinct fixed points on the Riemann sphere or one double fixed point. With two fixed points, conjugate them to zero and infinity. A Möbius map fixing both is $z\mapsto kz$ with $k\ne0$. With one fixed point, conjugate it to infinity; the resulting map is affine, $z\mapsto az+b$. If $a\ne1$ it would have another fixed point, so $a=1$ and $b\ne0$. Scaling the coordinate makes it $z\mapsto z+1$. The identity is already the case $k=1$.
Define $\operatorname{tr}^2(g)$ using a determinant-one representing <matrix>; the two possible lifts differ by sign, so its square is well defined and invariant under conjugation. For $z\mapsto kz$ it is $k+2+k^{-1}$, while for a nontrivial translation it is four. If two nonidentity dilations have equal trace-square, their multipliers satisfy the same quadratic and are either equal or reciprocal. Inversion $z\mapsto1/z$ conjugates one reciprocal multiplier to the other. Trace-square four forces the dilation multiplier to be one, which is the excluded identity; thus every nonidentity map with that invariant is in the translation class. This proves $\boxed{g\sim h\iff\operatorname{tr}^2(g)=\operatorname{tr}^2(h)}$ for the stated nonidentity maps.
\b[Not every such map fixes an interior point of hyperbolic three-space.] In its upper-half-space extension, $z\mapsto z+1$ translates the horizontal coordinate and has no fixed point at positive height. A dilation extends as $(z,t)\mapsto(kz,|k|t)$, so it fixes the vertical axis only when $|k|=1$; if $|k|\ne1$, the height changes. Boundary fixed points therefore do not imply an interior fixed point.
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