Solve the differential equation $f'=f/(f\circ f)$

I am completely stuck on the following differential equation:

Find all differentiable functions $ f:\mathbb{R}_{>0}\to\mathbb{R}_{>0} $ such that:$$ f' = \frac{f}{f\circ f}. $$

The identity function works, and if one assumes that $ f $ admits a fixed point $ \alpha > 0 $, then I even have an easy solution to show the identity function is the unique solution. It involves integrating from $ \alpha $ to any $ x > 0 $. This goes as follows:

Since $ f $ is continuous, we can define for any $x>0$:$$ F(x) := \int_{\alpha}^{x} f(t) \,\mathrm dt. $$

By the fundamental theorem of calculus, $ F:\mathbb{R}_{>0}\to\mathbb{R} $ is continuous. Moreover, since $ f $ is continuous, we also know that $ F $ is everywhere differentiable with $ F' = f $. Because $ f $ is always positive, $ F $ is strictly increasing and thus injective.

We now show that $ f = \operatorname{Id}_{\mathbb{R}_{>0}} $. Fix $ x \in \mathbb{R}_{>0} $. From the given equation (valid for all $t>0$),$$ f(t) = f(f(t)) f'(t) = F'(f(t)) f'(t) = (F\circ f)'(t), $$we obtain:$$ F(x) = \int_{\alpha}^{x} f(t) \,\mathrm dt = \int_{\alpha}^{x} (F\circ f)'(t) \,\mathrm dt = (F\circ f)(x) - (F\circ f)(\alpha), $$where we used the fundamental theorem of calculus again for the function $ x \mapsto \int_{\alpha}^{x} (F\circ f)'(t) \,\mathrm dt $ (it holds for both $ x \geq \alpha $ and $ x < \alpha $).

Now, using the fact that $ \alpha $ is a fixed point of $ f $, we get $ F(f(\alpha)) = F(\alpha) = 0 $, so that $ F(x) = F(f(x)) $. By injectivity of $F$, we conclude that $ f(x) = x $. Since $ x > 0 $ was arbitrary, we conclude.

I would greatly appreciate any help in finding either:

1. A counterexample showing that the identity function is not the only solution,
2. A proof that $ f $ must have a fixed point, or
3. An alternative proof of the claim.

Thanks!

Edit: A friend of mine originally received this question from someone else and passed it on to me. I saved it and, after two years, tried to solve it. When I asked my friend if he had found a solution, he told me he no longer remembered the question (great friend!). It is then possible that the problem is incomplete or that some details are missing.