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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 $f:[a,b]\to[a,b]$?

Tanuel Chi
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    I think that your exponent $\omega$ is wrong. It should be $\exp(\pm\frac{i\pi}3)$. – Denis Serre May 13 '19 at 13:14
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    https://mathoverflow.net/questions/111066/are-there-any-techniques-for-solving-a-differential-equation-of-the-form-f-x/111092#111092 I wonder why this question is so popular... – fedja May 13 '19 at 14:59

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