Apropos "fix"point: are the fractional iterations from there "fix" as well?

[Gottfried]

So, this is actually a defining property of the standard Schroder iteration. But it's a little difficult to fully flesh out why.  Now, to begin I'll construct an arbitrary iteration which has a constant noodle, and show there are many of them.
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Yes, my take of this has been: by the analysis of the power series including the conjugacy we know, that the powerseries for the Schroder-function has no constant term, and the original function \(f(z) \) at \( z=t \) (where \(t\) is the fixpoint), gives \( \sigma(t-t) = 0\) and is thus zero, so that any fractional iteration has \( \lambda^h \cdot \sigma(t-t) = \lambda^h \cdot 0 = 0 \) and by this \( \sigma°^{-1}(0) +t = t \)
is constant for every \(h \).
So the use of the Schroder-mechanism produces the constant \(t \) at each (integer or fractional or even complex) iteration height. (Interesting that you say it either the other way: if we want have the constant \(t\) at the fractional iteration then that implies that we'll need the Schroeder-mechanism, this looks good, but I'll have to chew on it.)

What is/was the unclear point for me is: is this an arbitrary decision to choose a model which has this constancy or are there more/deeper reasons, that the option for this model is required to keep things consistent.

You introduce then the property of being a semigroup, expressed by \( f°^a(f°^b(z)) = f°^{a+b}(z) \) - that would be something I would say, is demanded for any implementation of fractional iteration, so this would give a convincing argument that we do not have the freedom to choose the model, (it's not like with the axiom of the parallels in geometry, where we simply decide for another (legitimate) geometry if we decide for another version of the axiom).

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Superfunctions, don't have to have a constant noodle--but they don't have the semi-group property then.

I'd like this argument concerning the semigroup-property being valid, which you gave in the following, only at the moment I couldn't get it completely yet, I'll come back to this if I'm clearer with this.

Thanks -
Gottfried