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Work on the Riemann sphere, including zero and infinity. For $j(z)=1/z$, two iterations give the identity and one does not, so $\mathrm{ord}(j)=2$. For $k(z)=1/(1-z)$, direct composition givesNeither $k$ nor $k^2$ is the identity; in particular $0\mapsto 1\mapsto \infty \mapsto 0$. Thus

To prove the Möbius conjugacy normal forms, write a Möbius transformation as $f(z)=(az+b)/(cz+d)$ with $ad-bc\ne 0$. A finite fixed point satisfies $c{z}^2+(d-a)z-b=0$. If $c\ne 0$, this quadratic has two roots counting multiplicity, and infinity is not fixed. If $c=0$, the transformation is affine, infinity is fixed, and there is one further finite fixed point unless its slope is one. A nonidentity slope-one affine map is a nonzero translation with only infinity fixed. Thus any nonidentity Möbius transformation has one or two distinct fixed points.

If there are two, say $p,q$, choose a Möbius transformation $h$ taking them to zero and infinity. For finite $p,q$ one can use $h(z)=(z-p)/(z-q)$; the cases involving infinity use an affine map or reciprocal. The conjugate $hf{h}^{-1}$ fixes zero and infinity. Fixing infinity sets its denominator's linear coefficient to zero, and fixing zero sets its numerator's constant term to zero; hence it is $z\mapsto \mu z$ with $\mu \ne 0$.

If there is only one fixed point, send it to infinity. The conjugate has the affine form $az+b$. If $a\ne 1$, $b/(1-a)$ would be another fixed point, which is impossible; therefore $a=1,b\ne 0$. Conjugating this translation by $z\mapsto z/b$ yields $z\mapsto z+1$. The identity is already the scaling $\mu =1$. This proves the classification, and invertibility excludes $\mu =0$.

Conjugation carries fixed-point sets bijectively. The translation $z+1$ has exactly one fixed point, infinity. A nonidentity scaling has the two fixed points zero and infinity, while the identity fixes every point. Therefore $z+1$ is not conjugate to any scaling.

Finally let $f$ have finite positive order $n$. A nonzero translation has infinite order, since its $k$th iterate is $z+k$ after normalization. Thus $f$ is conjugate to $z\mapsto \mu z$, where $\mu$ has multiplicative order exactly $n$. For $n>1$, zero and infinity each form a singleton orbit. At a finite nonzero $z$, an iterate returns precisely when ${\mu }^{k}z=z$, equivalently ${\mu }^k=1$; the least positive return time is $n$. Conjugacy preserves orbit sizes. Hence the finite-order Möbius orbit sizes areFor $n>1$ exactly two points have size-one orbits; every other point has orbit size $n$.