$p(z) = 1-\frac{1}{z}$ has order 3: $p(p(p(z))) = z$
Is there an order 5 rational map with rational coefficients?
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Are there rational maps of order 4? (Order 2 is easy: $1/z$) – lhf Aug 03 '20 at 21:14
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@lhf $z\mapsto\frac{z+1}{-z+1}$? – Angina Seng Aug 03 '20 at 21:20
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Such a rational map would be a linear fractional transformation: $$f(z)=\frac{az+b}{cz+d}.$$ Its $k$-th iterate is the identity iff $A^k=\lambda I$ for some $\lambda$ where $A=\pmatrix{a&b\\c&d}$. If the coefficients $a,\ldots,d$ are rational this is impossible for $k=5$ unless $f$ is already the identity. The eigenvalues of $A$ would have to be $\sqrt[5]{\lambda}\zeta$ and $\sqrt[5]{\lambda}\zeta^{-1}$ for some non-trivial fifth root of unity $\zeta$. Their sum and product must be rational, so that $\lambda^{2/5}\in\Bbb Q$ which means that $\sqrt[5]{\lambda}$ is rational. But then $\sqrt[5]{\lambda}(\zeta+\zeta^{-1})$ is irrational, a contradiction.
Angina Seng
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