Fibonacci as iteration of fractional linear function

[bo198214]

Well bo, if you take abelian property, it'll fail then since this function can be conjugated by diagonal: \(\phi=\frac{\sqrt{5}-1}{2}\)
\(f(z)=\frac{1}{1+z},g(z)=\frac{\phi z-\phi-1}{z+1},g^{-1}(f(g(z)))=-(\phi+2)z\)
When you're asking a real valued super function of \(f(z)=\frac{1}{1+z}\)
You're actually transforming a super function problem to another, the relation between the superfunctions of \(f(z)\) and \(s(z)=-(\phi+2)z\) is
\(\alpha^{-1}\{f\}(z)=g(\alpha^{-1}\{s\}(z))\)
And since \(g:\mathbb{R}\to\mathbb{R}\), to find a function \(\alpha^{-1}\{f\}:\mathbb{R}\to\mathbb{R}\), you've equitably to find a real-valued superfunction of \(s(z)\approx-2.618z\)
That is to solve \(R:\mathbb{R}\to\mathbb{R},R(z+1)=-2.618R(z)\)
And as you take abelian property, if you assume \(s^t(z)=R(t+R^{-1}(z))\) you have to solve
\(R:\mathbb{R}\to\mathbb{R},R(z+1)=-2.618R(z)\) for a monotonic R 

Seems really we have to find a common language here:
So what is \(\alpha^{-1}\{f\}(z)\) what is this braces notation?
What is the Abelian property?

If it's continuous then it has to be a constant... so no solution if allowing the superfunction to be invertible

Yeah that was the thing with JmsNxn's superfunction at eta minor: It was not invertible, rather close to periodic.