Inverse super-composition

[sheldonison]

Your question is too general, since you don't identify what f(x) you are interested in. In general, the type of solution depends on the behavior at the fixed point. I assume you are interested in real valued functions. Some iterated functions have an attracting point. Then we look at the slope at the fixed point.

Yes, my question is general, because I am looking for a totally general solution for all the kind of problem like this.
Anyway, I found a nicer formula for my question instead of steinix-ankh, like this:
\( f(x) ^ o ^N = g(x) \)
According to the knowledge of f and g, what is N? How can I calculate it?
For example:
\( 2x ^ o ^N = x^2 \)
Thus N must be = log2(x), but here is the question why and how can I know it from?

For a fixed point of zero, with a fixed point multiplier of 2, the general solution for the Abel function generated at the fixed point of zero is:
\( f(z) = 2x + \sum_{n=2}^{\infty}a_n z^n \)

\( \alpha(z) = \log_2(S(z))\;\;\; \) This is the Abel function for f(z) \( \;\alpha(f(z)) = \alpha(z)+1 \)

where S(z) is the formal Schröder equation solution;
\( S(f(z)) = 2\cdot S(z)\;\;\; S(z)=z+\sum_{n=2}^{\infty}b_n z^n\;\; \)
This is sometimes called Koenig's solution. It can be modified to work with any fixed point multiplier of k, |k|<>1. Using pari-gp one can easily write a program to generate the formal power series for S(x) given f(x).