How many functions are there which are differentiable on $(0,\infty)$ and they satisfy the relation $f^{1}=f'$.

Let $a=1+p>1$ be given. We shall construct a function $f$ of the required kind with $f(a)=a$ by means of an auxiliary function $h$, defined in the neighborhood of $t=0$ and coupled to $f$ via $x=h(t)$, $f(x)=h(a t)$, $f^{1}(x)=h(t/a)$. The condition $f'=f^{1}$ implies that $h$ satisfies the functional equation $$(*)\quad h(t/a) h'(t)=a h'(at).$$ Writing $h(t)=a+\sum_{k \ge 1} c_k t^k$ we obtain from $(*)$ a recursion formula for the $c_k$, and one can show that $0< c_r<1/p^{r1}$ for all $r\ge 1$. This means that $h$ is in fact analytic for $t< p$, satisfies $(*)$ and possesses an inverse $h^{1}$ in the neighborhood of $t=0$. It follows that the function $f(x):=h(ah^{1}(x))$ has the required properties. 


Wow. I remember that I thought exactly the same problem out of curiosity as a high school student but did not reach an answer. In fact, I was thinking about posting this problem on MathOverflow! At least it is easy to construct one solution: f(x)=x^{φ}/φ^{φ−1}, where φ=(1+√5)/2 is the golden ratio. Edit: Corrected the calculation. Thanks to Aaron Meyerowitz for spotting the error! 

