2 real fixpoints again .......

[sheldonison]

Consider a function f that is strictly increasing , u-shaped and real-analytic for real > 0.

With fixpoints at 0 with derivative 0 and fixpoint at 1 with derivative 2.

Having a derivitive=0 at x=0 is the same as a super-attracting fixed point at x=0. The Op is refering to the classic example:
\( f(x)=x^2 \)
whose superfunction is the same from both fixed points. The fixed point at 1 is repelling with a period of 2, and there is also a super attracting fixed point at infinity.
\( f^{\circ z}(0.5) = 2^{(-2^z)} \)

For f(x)=x^2, this is the same as the Schroeder function solution from the fixed point of 1.

For all other super-attracting x^2 functions, \( f(x)=x^2+\sum_{n=3}^{\infty}a_n\cdot x^n\; \) one can probably prove that the two superfunctions, one from the superattracting fixed point, and the other from the repelling fixed point, will always be different. One could look at Miller's book for the general form for putting the super-attracting fixed point function in congruence with the unit circle. There will be a fractal Julia set for the basin of attraction for the super-attracting fixed point, and the repelling fixed point of 1 will be on the Julia set boundary. We can always take the Abel function of the other fixed point's Superfunction, to generate a theta mapping. But theta mappings give us an infinite number of half iterates, abel functions, and super functions possible, all valid, besides the two formal solutions at the fixed points. It sort of reminds me of the sexp/slog sqrt(2) case, except without a uniqueness criteria. The sqrt(2) case has a uniqueness criteria when viewed as an extension of tetration.