Also; I must've missed the challenge to construct an inverse abel function. I'll be hard pressed if the following method doesn't work. If it doesn't, I'll try another.
Call \( \mathcal{P} \) the set in which,
\(
f(\xi) = -\xi(1-\xi)\\
\lim_{n\to\infty} f^{\circ n}(\xi) = 0
\)
Take a neighborhood of zero \( \mathcal{U} \) and intersect with \( \mathcal{P} \) to get \( \mathbb{E} = \mathcal{U} \cap \mathcal{P} \). Change our variables to \( g(\xi) = f^{\circ 2}(\xi) \). Then, for some \( T \in \mathbb{R}^+ \),
\(
f^{\circ 2z}(\xi) : \mathbb{C}_{\Re(z) > 0,\Im(z) < T} \times \mathbb{E} \to \mathbb{E}\\
f^{\circ 2z}(\xi) = \frac{d^{z-1}}{dw^{z-1}}|_{w=0} \sum_{n=0}^\infty f^{\circ 2(n+1)}(\xi)\frac{w^n}{n!}\\
\)
Now, if we take the fourth root we will get \( h(\xi) = f^{1/2}(\xi) : \mathbb{E} \to \mathbb{E} \); but, you're correct; you'll get \( h'(0) = 1 \)--when we want \( h'(0) = \pm i \). The trick is easier than you're making it. As this converges, let's create a family \( f_\lambda^{\circ z} \) for \( 0 < \lambda < 1 \) where \( \lambda \to 1 \) in the limit and gives us the above iteration; \( \frac{d}{d\xi}|_{\xi =0} f_\lambda^{\circ 2z}(\xi) = \lambda^{2z} \).
To begin,
\(
f_\lambda^{\circ 2z(1+\pi i k/\log|\lambda|)}(\xi)
\)
Is also an iteration, but,
\(
\frac{d}{d\xi}|_{\xi = 0} f_\lambda^{\circ 2z(1+\pi ik/\log|\lambda|)} = e^{2z \log(\lambda)+2\pi i k z}
\)
Now, when we take the fourth root there are four options, one negative and one positive; and two imaginary. And voila; if we limit \( \lambda \to 1 \) we've avoided your problem and we find that through a limit process we can make a square root function which satisfies,
\(
h'(0) = \pm i\\
\)
I hope that answers your question; if not I'll go into more detail. I only know how to compute these things through limits; but they again, relate to Ecalle. They might display odd complex behaviour though; or cease to exist at certain points. I tend to find a good heuristic, is to take a limit \( \lambda f \) as \( \lambda \to 1 \); to discover how to control iterates; and choose different iterates. Remember; there are at most n functions about a fixedpoint which are n roots of a function. I've yet to find a function that didn't have exactly n roots about that fixed point though.
Call \( \mathcal{P} \) the set in which,
\(
f(\xi) = -\xi(1-\xi)\\
\lim_{n\to\infty} f^{\circ n}(\xi) = 0
\)
Take a neighborhood of zero \( \mathcal{U} \) and intersect with \( \mathcal{P} \) to get \( \mathbb{E} = \mathcal{U} \cap \mathcal{P} \). Change our variables to \( g(\xi) = f^{\circ 2}(\xi) \). Then, for some \( T \in \mathbb{R}^+ \),
\(
f^{\circ 2z}(\xi) : \mathbb{C}_{\Re(z) > 0,\Im(z) < T} \times \mathbb{E} \to \mathbb{E}\\
f^{\circ 2z}(\xi) = \frac{d^{z-1}}{dw^{z-1}}|_{w=0} \sum_{n=0}^\infty f^{\circ 2(n+1)}(\xi)\frac{w^n}{n!}\\
\)
Now, if we take the fourth root we will get \( h(\xi) = f^{1/2}(\xi) : \mathbb{E} \to \mathbb{E} \); but, you're correct; you'll get \( h'(0) = 1 \)--when we want \( h'(0) = \pm i \). The trick is easier than you're making it. As this converges, let's create a family \( f_\lambda^{\circ z} \) for \( 0 < \lambda < 1 \) where \( \lambda \to 1 \) in the limit and gives us the above iteration; \( \frac{d}{d\xi}|_{\xi =0} f_\lambda^{\circ 2z}(\xi) = \lambda^{2z} \).
To begin,
\(
f_\lambda^{\circ 2z(1+\pi i k/\log|\lambda|)}(\xi)
\)
Is also an iteration, but,
\(
\frac{d}{d\xi}|_{\xi = 0} f_\lambda^{\circ 2z(1+\pi ik/\log|\lambda|)} = e^{2z \log(\lambda)+2\pi i k z}
\)
Now, when we take the fourth root there are four options, one negative and one positive; and two imaginary. And voila; if we limit \( \lambda \to 1 \) we've avoided your problem and we find that through a limit process we can make a square root function which satisfies,
\(
h'(0) = \pm i\\
\)
I hope that answers your question; if not I'll go into more detail. I only know how to compute these things through limits; but they again, relate to Ecalle. They might display odd complex behaviour though; or cease to exist at certain points. I tend to find a good heuristic, is to take a limit \( \lambda f \) as \( \lambda \to 1 \); to discover how to control iterates; and choose different iterates. Remember; there are at most n functions about a fixedpoint which are n roots of a function. I've yet to find a function that didn't have exactly n roots about that fixed point though.