## Find real function $f(x)$ such that $f(f(x))=f'(x)$

Absolutely there is a trivial solution $$f(x)=0$$. Actually, assuming $$f(x)$$ being smooth and expanding $$f(x)$$ into power series one can get $$f(0)=0\to f(x)=0$$. Also, in the complex field there are solutions e.g. $$f(x)=(-\omega)^{\left(-\omega^2\right)}x^{-ω}$$ where $$\omega=e^{\pm 2\pi i/3}$$. So I am wondering, are there any non-zero $$f:\mathbb{R}\to \mathbb{R}$$ such that $$f(f(x))=f'(x)$$?

If there is no solution on $$\mathbb{R}$$, then are there any solutions on some real interval, such as $$f:[a,b]\to[a,b]$$?