Find all holomorphic functions $f$ (on $\mathbb{C}$) satisfying $f(1-f(z))=f(z)$?
First, every constant function $f(z)=w$ is holomorphic and satsfies our condition. Now assume $f$ is not constant. So there is some $z_0$ such that $f'(z_0)\ne 0$.
Every $y$ in range $f$ must satisfy $f(y)=1-y$. Now, since range of $f$ has non-empty interior and is entire, $f(z)=1-z$ everywhere.
Is that correct? I have some feeling Im missing sth...
Yes, OP is right. One can easily redefine OP's problem ($f \leftrightarrow 1-f$) as asking the following.
Find all holomorphic functions $f$ (on $\mathbb{C}$) satisfying $$\tag{1} f\circ f~=~f.$$
Possible elementary method: Use the chain rule
$$\tag{2} (f\circ f)^{\prime}(z)~=~ f^{\prime}(f(z))f^{\prime}(z) ~=~f^{\prime}(z). $$
We already know that the constant functions $f$ are solutions to (1), so assume that $f$ is a non-constant function. Hence we can assume that exists a point $a\in\mathbb{C}$ (and an open neighborhood $U$ of $a$) so that $f^{\prime}$ does not vanish in $U$. Then eq. (2) means (using the inverse function theorem) that there exists an open neighborhood $V$ of $f(a)$ so that
$$\tag{3} \forall w\in V:~~ f^{\prime}(w)~=~1. $$
Eq. (3) implies that there exists an integration constant $b$ so that $$\tag{4} f(w)~=~w+b.$$ Plugging eq. (4) back into eq. (1) yields $b=0$. So $f$ is the identity map.
