My question is an amalgamation of two previous questions. The first question I'd like to draw attention to is here. It asks whether there can exist a non trivial semigroup defined on $\mathbb{C}$

$$\phi(s,z):\mathbb{C}_{\Re(s) > 0}\times\mathbb{C} \to \mathbb{C}$$ $$\phi(s_0, \phi(s_1,z)) = \phi(s_0+s_1,z)$$

The answer is no, because such a semigroup would necessarily have a repelling fixed point for some $s_0$, and each function commutes with each other, and an entire function with a repelling fixed point can only have a countable number of functions that commute with it; which forms a contradiction. A big thanks goes to Alexandre Eremenko for referencing I.N. Baker and solving this.

The second question is found here and asks whether an entire function $g$ of finite order can have a composite square-root. The answer is yes. Again, another big thanks goes to Alexandre Eremenko for referencing I.N. Baker again, and solving this.

A great question which hybridizes the above two questions, but laying insoluble in both instances:

For an entire function $f:\mathbb{C} \to \mathbb{C}$, can there exist a sequence of entire functions $\{f_n\}_{n=1}^\infty$ such that: $$f_n(f_n(...(n\,times)...f_n(z))) = f_n^{\circ n}(z) = (f_n \circ f_n \circ ...(n\,times)...\circ f_n)(z) = f(z)$$

This first answer fails to prove this can't happen, this is only a countable list of functions that commute with $f$, so the proof provided by Baker does not work. The second answer suggests this might be possible, but only refers to a square root of $f$, not an arbitrary $n$'th root for all $n$.

Essentially I'm asking this because I've learnt that if $f:\mathbb{C} \to \mathbb{C}$ there is no valid way of constructing

$$(f \circ f \circ...(s\,times)...\circ f)(z) : \mathbb{C}_{\Re(s) > 0}\times\mathbb{C} \to \mathbb{C}$$

so I'm wondering if we can weaken this to, sure there is no semigroup, but there can be

$$(f\circ f\circ... (q\,times)...\circ f)(z):\mathbb{Q} \times \mathbb{C} \to \mathbb{C}$$

My money is on the fact this is impossible and I'm crossing my fingers that Alexandre Eremenko knows why because he's read so much Baker.