Given any $n\in \mathbb{N}$, find an analytic function $f$ with the properties that
(i) $f$ composed with itself $n$ times gives $f^{\circ n}(x) = x$ and
(ii)$f$ satisfies $f(0) = 0$ and $f'(0) = 1.$
For some inspiration, some functions that satisfy $f^{\circ n} = f$ include $$f(x) = \tan (\tan^{-1}(x) + \frac{\pi}{n})$$ and $$f(x) = \frac{x\cos\frac{\pi}{n} - \sin\frac{\pi}{n}}{x\sin\frac{\pi}{n} + \cos\frac{\pi}{n}}.$$
I would really appreciate any help on this problem.
Note that $f(z)=z$ is a solution. Now let $f$ any solution. If $f(z)\not =z$, we can write $f(z)=z+az^m+..$ with $a\not =0$ and $m\geq 2$. We compute: $$f(f(z))=f(z)+af(z)^m+...=z+2az^m+..$$
By an easy induction, we get that for $k\geq 1$, we have $$f^{ok}(z)=z+kaz^m+...$$
For $k=n$, we get that $ka=0$, a contradiction as $a\not =0$. Hence the only solution is $f(z)=z$.