A composite square root of a function $g$ is a function $f$ such that $f(f(z)) = g(z)$. Not surprisingly, for arbitrary $g$ a function like this is hard to find. Specifically I am looking at functions $g$ such that $g$ has finite nonzero order, so that

$$0 < \limsup_{r\to\infty} \dfrac{\log\log M(r)}{\log r} = \rho < \infty$$

where $M(r)$ is the max of $g(z)$ on the disk of radius $r$.

Now naturally $\cos(\cos(z))$ has a square root function $\cos$, or $e^{e^z}$ has one, but these functions are of infinite order. Similarly with polynomials, any polynomial $p(p(z))$ has a square root function, but these are of zero order. The special case of functions of finite order, is that a possible square root function is of zero order. Which is an argument simple enough (I'll provide it if asked).

The evidence I am seeing piling is that there exists no entire composite square roots of functions $g$ of finite order. Now if $I$ is the immediate basin about a geometrically attracting fixed point of $g$ (and $I$ is simply connected) then there is an $f$; but $f$ only sends $I \to I$ and is not entire. I'm wondering if this is always the case; that a composite square root cannot be entire.

Taking $g(z)=e^z$ for example, if there were such an $f$, then $f(f(z))$ can't equal zero, so that $f(z)$ misses $f^{-1}(0)$ but then $f(f(z))$ must miss $0$ and $f^{-1}(0)$ and must therefore be constant by Picard's theorem. If $f^{-1}(0)$ is empty than $f(f(z)) = e^z$ misses $0$ and $f(0)$. This argument generalizes for all $e^{P(z)}$ where $P$ is a polynomial.

Therefore the only possible cases where there is an $f$ for arbitrary $g$ are when $g$ is surjective on $\mathbb{C}$. This doesn't work for $\cos$ though, because $\cos$ has all its zeroes in $I$, the immediate basin about $z_0 \in \mathbb{R}^+\,\,\, z_0 \approx 0.739085$. It can be shown that if $f$ were entire all it's zeroes are described by the sequence $\{f(\pi/2 + n\pi)\}_{n=-\infty}^\infty$, but $f$ is necessarily periodic (again an argument I will supply if asked) so that $f$ would only have two zeroes $\pi/2$ and $3\pi/2$ and thus by Hadamard's factorization theorem $f(z) = C(z-f(\pi/2))(z-f(3\pi/2))$ which is obviously wrong. Thus no entire square root of $\cos$. The same argument applies for $\sin$ just as well but is a bit more finicky.

As much as I look, the only possible candidates are when we have an entire function $f$ of zero order and we compose it with itself $f(f(z)) = g$ and $g$ is of finite nonzero order. I've yet to come across an example of such a function though. Can anyone even name such a function?

So all in all my question is rather simple. For an entire function $g$ of finite nonzero order, can there exist an entire function $f$ such that $f(f(z)) = g(z)$?