If $f$,$g$ are entire functions such that and $g′(z)=f(g(z))$ everywhere, then show that $f$ is linear or $g$ is constant.
This claim was mentioned without proof in this related question (Applications of the Little and Great Theorems of Picard). I am unable to prove this.
One way might be to show that $f$ is injective by using Picard's Little Theorem - but I have no idea how to proceed.
Assume that there is such a solution with $g(a)=g(b)$. Then $h(z)=g(z-a+b)$ is such that $h(a)=g(a)$ and $h’=f(h), g’=f(g)$, thus $h=g$ so $g$ is $b-a$-periodic.
In particular, we can assume that $g$ is linear or that its set of periods is $2i\pi\mathbb{Z}$. In the first case, $f$ is constant (unless $g$ is).
In the second case, $g(z)=h(e^z)$, and $h$ is injective. But $h$ has singularities at $0$ and $\infty$, thus $h$ is meromorphic on $\mathbb{CP}^1$ (by eg Casorati-Weierstrass, or Picard’s theorem) and injective on that projective line minus two points, so it is a rational fraction of degree $1$ defined on $\mathbb{C}^{\times}$, so that $g(z)=C+De^{\pm z}$, and $g’(z)=\pm D e^{\pm z}$, so that $f(t)=\pm (t-C)$ is linear.