I was looking at the equation $f^{-1}(x)=\int f(x)dx$ recently.
One can note that it has an easy real-valued solution $f(x)=\phi^{\frac{\phi-1}{\phi}}x^{\phi-1}$ (by guessing for a solution of the form $ax^b$), where $\phi$ is the golden ratio, since:
$$f^{-1}(x)=\Bigg(\frac{x}{\phi^{\frac{\phi-1}{\phi}}}\Bigg)^{\frac{1}{\phi-1}}=\frac{x^{\frac{1}{\phi-1}}}{\phi^{\frac{1}{\phi}}}=\frac{x^{\frac{\phi}{\phi^2-\phi}}}{\phi^{\frac{1}{\phi}}}=\frac{x^{\phi}}{\phi^{\frac{1}{\phi}}}=\frac{\phi x^{\phi}}{\phi \phi^{\frac{1}{\phi}}}=\frac{\phi^{\frac{\phi-1}{\phi}} x^{\phi}}{\phi}=\frac{\phi^{\frac{\phi-1}{\phi}} x^{(\phi-1)+1}}{(\phi-1)+1}\\=\int f(x)dx$$
But, initially, I tried to avoid making such outright guesses.
I assumed a real analytic solution existed- then, I converted the equation into a slightly friendlier form:
$$f'(x)=\frac{1}{f(f(x))}$$
Then, finally, I assumed the analytic solution had a fixed point, $a$.
This would allow me to compute the function's derivatives about this, as-yet-unknown fixed point $a$ in terms of $a$ by repeatedly differentiating both sides of $f'(x)=\frac{1}{f(f(x))}$ (which I meant to plug into the Taylor series for the solution).
For instance, $f'(a)=\frac{1}{f(f(a))}=\frac{1}{a}$. By differentiating both sides of $f'(x)=\frac{1}{f(f(x))}$, we get $f^{(2)}(x)=\frac{-f'(f(x))f'(x)}{f(f(x))^2}$ which can be rewritten as $f^{(2)}(x)=\frac{-1}{f(f(x))^3f(f(f(x)))}$ to give $f^{(2)}(a)=\frac{-1}{a^4}$.
Similarly, the expressions for $f^{(3)}(a),f^{(4)}(a),$ and $f^{(5)}(a)$ are:
$$\frac{3}{a^7}+\frac{1}{a^8},\quad -\frac{15}{a^{10}}-\frac{10}{a^{11}}-\frac{3}{a^{12}}-\frac{1}{a^{13}},\quad \frac{105}{a^{13}}+\frac{105}{a^{14}}+\frac{55}{a^{15}}+\frac{30}{a^{16}}+\frac{10}{a^{17}}+\frac{3}{a^{18}}+\frac{1}{a^{19}}$$
and so on.
What I found interesting, coming back to the simple solution now, is that $f(x)=\phi^{\frac{\phi-1}{\phi}}x^{\phi-1}$ does in fact have a fixed point- namely, $x=\phi$.
It's easy to see that $f^{(k)}(\phi)=\phi^{1-k}\prod_{i=1}^k(\phi-i)$, so $f^{(k)}(\phi)$ should equal $$\sum_{j=-1}^{k-1}a_j\phi^{-j}$$ where
$$a_j=(-1)^{j+1}\sum_{S\subset \{1,2,...,k\},\\|S|=j+1}\prod_{x\in S} x$$
But, plugging in $\phi$ in place of $a$ (the fixed point), we get an expression of a similar form to $\sum_{j=-1}^{k-1}a_j\phi^{-j}$ for $f^{(k)}(\phi)$ but with different bounds and coefficients (even though both evaluate to the same thing). I wrote a simple script to calculate some of the higher derivative expressions for $a$ and using the numbers it gave me I found this OEIS sequence (it seems to match the coefficients of the expressions aside from the signs).
My question is, first, obviously, how can I determine whether the original functional equation has any other analytic solutions (in particular, ones with fixed points)? But, second, how can I prove the form for the coefficients of the derivative expressions in $a$ and what does it have to do with $\phi$?
Edit: If it helps, here's the script (in Python). If you spot anything wrong, please let me know
