Let $f$ be an entire function such that $f\circ f$ has no fixed points. Prove that $f$ is a translation $$z\mapsto f(z)=z+b \qquad (b\neq 0)$$
Firstly, we prove that there exists a constant $c\in \mathbb{C}\backslash \{0,1\}$ such that $$f(f(z))-z=c(f(z)-z) $$ applying Picard's little theorem. If $c=0$, then $f(f(z))=z$, so $f\circ f$ has a fixed point (absurd). If $c=1$, then $f(f(z))=f(z)$, so $f$ is the identity $f(z)=z$ and of course it has fixed point (absurd). Then, $$F(z)=\frac{f(f(z))-z}{(f(z)-z)}$$ is an entire function which does not take the values 0 and 1 so, by Picard's little theorem, it must be constant.
Also, I've proved that $f'\circ f$ is a constant function. Let's see this. Differentiating $$f(f(z))-z=c(f(z)-z) $$ we have $$f'(z)f'(f(z))-1=cf'(z)-c$$ $$f'(z)[f'(f(z))-c]=1-c$$ Again, the entire function $$G(z)=f'(z)[f'(f(z))-c]$$ does not take the values $0$ and $1$ so, by Picard's Little Theorem, then is constant.
However, I don't know how to prove this problem. Any help would be appreciate.
Note there may or may not be considerable overlap between what's below and what's in the current version of the question; if so it's because the OP was revising the question while I was typing the answer. What actually happened: In the original version of the question he or she asserted that $f'\circ f$ was constant. I showed how the result followed from that and asked how to show that $f'\circ f$ was constant. He gave a few not quite right arguments to that effect while I was concocting my own proof. He gets credit for $f(f(z))-z=c(f(z)-z)$ and for saying that it follows that $f'\circ f$ is constant...
Hmm. Took me a minute to see how to prove the first thing you say you've proved. I don't see yet how to show that $f'\circ f$ is constant. But if that's correct you're done: Since $f$ has no fixed point $f$ is nonconstant; so the range of $f$ is dense, hence $f'(f(z))=k$ for all $z$ implies $f'(z)=k$ for all $z$.
How do you show $f'\circ f$ is constant?
Ah, here's how you show that: First, if $c=0$ then $f\circ f$ has lots of fixed points. So $c\ne0$. Differentiating the first identity shows that $$f'(z)(f'(f(z))-c)=1-c.$$
Hence $f'\circ f$ cannot take the values $0$ or $c$. If $f'\circ f(w)=c$ then $c=1$, so $f(f(z))=f(z)$, for all $z$,, so $f(z)$ is a fixed point of $f$ and hence of $f\circ f$. And if $f'\circ f(w)=0$ then in particular $f'$ has a zero, so it follows again that $c=1$.
So Picard shows that $f'\circ f$ is constant (since $c\ne0$).
For the benefit of anyone confused by the proof in the OP that $f(f(z))-z=c(f(z)-z)$: since $f$ has no fixed point, $$F(z)=\frac{f(f(z))-z}{f(z)-z}$$ is entire. If $F(z)=0$ then $f\circ f$ has a fixed point,, while $F(z)=1$ implies $f(f(z))=f(z)$, so $f$ has a fixed point. So Picard shows $F$ is constant.