= Solution
For $f(z)=(az+b)/(cz+d)$, finite fixed points obey
$$
cz^2+(d-a)z-b=0,
$$
with the point at infinity included in the usual way when appropriate. The fundamental theorem of algebra on the Riemann sphere gives at least one fixed point. Unless $f$ is the identity, the equation is nonzero of degree at most two, so there are one or two distinct fixed points.
Let $\zeta$ be a primitive $m$th root of unity. For every $u\in\mathbb C$,
$$
f_u(z)=u+\zeta(z-u)
$$
has order $m$, and these transformations are distinct as $u$ varies.
Projectivising <matrices> is a homomorphism. Thus $B=CAC^{-1}$ in $SL_2(\mathbb C)$ implies $g=[C]f[C]^{-1}$ in the Möbius <group>.
The converse as stated is false because <matrix> representatives may be rescaled: $A=I$ and $B=2I$ define the same Möbius transformation, while neither $B$ nor $-B$ is conjugate to $A$ (their traces differ).
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