Iteration with two analytic fixed points

[bo198214]

So this seems to be a dead end, it is as frustrating for me.
Ok, try it in a different way. You say:

can we agree that if \(f^{\circ t}(z) : D \times H \to H\) for two domains \(D\subset \mathbb{C}\) and \(H\subset \mathbb{C}\)--where \(D\) is closed under addition (is a semigroup under \(\{+\}\)); such that:

\[
\begin{align}
f^{\circ t}(f^{\circ s}(z)) &= f^{\circ t+s}(z)\\
f^{\circ 1}(z) &= f(z)\\
\end{align}
\]

Then there can only be one point \(p \in H\) such that \(f(p) = p\).

But for me this has nothing to do with non-integer iteration. Your proposition would already be valid for just (positive) integer iteration, isn't it? (I rephrase your proposition to see what I mean)

if \(f^{\circ n}(z) : \mathbb{N} \times H \to H\) for a domain \(H\subset \mathbb{C}\):

Then there can only be one point \(p \in H\) such that \(f(p) = p\).

I didn't think deeply about it. It could be wrong. If you have a counter-example please put it out.

But if this is true, then the preconditions of your statement are so restrictive that just no function with two fixed points can fit.
And then you can not call it a statement about non-integer iteration.

I'm sorry bo, but for you to say I am the one repeating myself isn't correct.

This is already the second time that you claim I said something that I didn't say (and also didnt mean).
Please be more careful what you allege people have said.
I don't know how one can misread this:

I thought I emphasized that too much already, and now you ask me whether I understand that?