Functions satsify $f'=f^{-1}$ with $f^{-1}$ is compostional inverse of$ f$

Question

let $f$ be a function such that :$f:\mathbb{C}\to \mathbb{C}$ and $f^{-1}$ is the compositional inverse of $f$, I seek for the analyticity of $f$ at $0$, then my question here is :

Question: Are there functions satisfy:$f'=f^{-1}$ with $f^{-1}$ is compostional inverse of $f$ ?

Answer 1

Such a function appears to not exist. If $f^{-1}$ exists, then $f^{-1}(f(0))=0$. If $f^{-1}=f'$, then $f'(f(0))=0$, which implies that $f$ is not invertible in a neighborhood of $f(0)$.

Answer 2

If $f'=f^{-1}$ and $f(x) = ax^b$, then $f'(x) = ab x^{b-1}$ and $f^{(-1)}(x) =(x/a)^{1/b} $ so $(x/a)^{1/b} =ab x^{b-1} $. If $b \ne 0$, then, raising to the $b$ power, $x/a =(ab)^b x^{b(b-1)} $ or $a^{-b-1}b^{-b} = x^{b(b-1)-1} $.

For the right side to be constant, we must have $0 =b(b-1)-1 =b^2-b-1 $ so $b =\dfrac{1\pm \sqrt{1+4}}{2} =\dfrac{1\pm \sqrt{5}}{2} $ and $a^{-b-1}b^{-b} = 1 $ or $a^{-b-1} = b^b $ or, since $b+1 = b^2$, $a =b^{\frac{b}{-b-1}} =b^{\frac{b}{-b^2}} =b^{\frac{-1}{b}} $.