Fibonacci as iteration of fractional linear function

[Leo.W]

However I wonder why it seems to be regular at the right fixed point ...

It's because it is not an iteration at all. It satisfies \(f^{\circ t+1} = \frac{1}{1+f^{\circ t}} \) (which corresponds to the Fibonacci identity) but it does not satisfy \(f^{\circ s+t}=f^{\circ s}\circ f^{\circ t}\), while the original Fibonacci extension is a true (and regular) iteration.
So the question still stands to find a real valued super function \(\frac{1}{1+z}\).

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 
If it's continuous then it has to be a constant... so no solution if allowing the superfunction to be invertible
If not assumed \(s^t(z)=R(t+R^{-1}(z))\), you still have to work out an abelian iteration mapping R to R for s(z) lol