What is the functional inverse (with respect to $h$ (!)) of $f^{\circ h}(x)={F(h) +x F(h-1) \over F(1+h) +x F(h) }$? I've considered the fractional iteration of $f(x) = {1 \over 1+x} $ for which the general expression depending on the iteration-height parameter $h$ might be assumed by the formula 
$$ f^{\circ h}(x) = f(x,h)={F(h) +x \cdot F(h-1) \over F(1+h) +x \cdot F(h)  }$$  where
$ \displaystyle \qquad \small F(h)=\operatorname{fibonacci}(h) = {\varphi^h - (1-\varphi)^h \over \sqrt5 }$ and      
$ \displaystyle \qquad \small \varphi = {1+\sqrt5 \over 2} \sim 1.618... $        
(Note, that for fractional iterates we need $x \in \mathbb C$) 
Today I tried to find a formulation for the functional inverse with respect to h, but don't find a good starting point so:    
Q: How would look a function $ h = \operatorname{hgh}(x_0,x_h) $ which would indicate the required iteration-height $h$ given $x_0$ and $x_h = f(x_0,h)$ ? Is there even a closed form for it (I'd call it still closed-form  if it possibly includes e.g. the Lambert-W function)  
[update]: for more background see the older exercise of mine
 A: Fractional linear transformation $f$ can be represented by the $2\times 2$ matrix $\displaystyle \left(\begin{array}{cc} 0 & 1 \\ 1 & 1\end{array}\right)$, which is diagonalized as
$$\left(\begin{array}{cc} 0 & 1 \\ 1 & 1\end{array}\right)=
P\left(\begin{array}{cc} \varphi & 0 \\ 0 & 1-\varphi\end{array}\right)P^{-1},\qquad 
P=\left(\begin{array}{cc} \varphi-1 & -\varphi \\ 1 & 1\end{array}\right).$$
Therefore the matrix corresponding to $h$th iteration is
$$P\left(\begin{array}{cc} \varphi^h & 0 \\ 0 & (1-\varphi)^h\end{array}\right)P^{-1}=
\frac{1}{\sqrt{5}}\left(\begin{array}{cc} (\varphi-1)\cdot\varphi^h +\varphi\cdot(1-\varphi)^h & \varphi^h -(1-\varphi)^h \\ \varphi^h -(1-\varphi)^h & 
\varphi\cdot\varphi^h +(\varphi-1)\cdot(1-\varphi)^h \end{array}\right)$$
and we obtain
$$\left(\frac{1-\varphi}{\varphi}\right)^{h}=\frac{x_hx_0+\varphi x_h-(\varphi-1)x_0-1}{x_hx_0-(\varphi-1)x_h+\varphi x_0-1}.$$
Taking the logarithm, one finds the expression
$$\mathrm{hgh}(x_h,x_0)=\ln\frac{x_hx_0+\varphi x_h-(\varphi-1)x_0-1}{x_hx_0-(\varphi-1)x_h+\varphi x_0-1}\Bigl/\Bigr.\ln\frac{1-\varphi}{\varphi}.$$
Note, however, that the expressions in the logarithms can become negative or even complex (in the denominator logarithm it is always negative), so some care should be taken to choose correct branches.
A: Ahh, the hints to the matrices and to the superfunction were useful. They forced me to look more seriously in my own exercise - and of course the Abel-function is the log of the Schröder-function and in the exercise I've already defined a very simple series for the Schröder-function.     
Let's denote $ \small \beta = \varphi-1 \sim 0.618...$ and $ \small u =-\beta^2$ then according to my own formula 2.7 the (recentered) Schröder function is
$$ \sigma (t) = -\sqrt5 \sum_{k=1}^\infty (-{t \over \sqrt5})^k $$ and the inverse schröderfunction 
$$ \sigma^{-1} (t) = \sqrt5 \sum_{k=1}^\infty ({t \over \sqrt5})^k $$ 
such that the continuous iterate is
$$ f(x,h) = \sigma^{-1} \left(u^h \cdot \sigma( x - \beta) \right) + \beta $$
The height-function $hgh(x_h,x_0)$ can be taken from the logarithm to the base $u$ of the ratio of the schröder-values as
$$ \operatorname{hgh}(x_h,x_0) = \log_u \left({ \sigma(x_h - \beta) \over \sigma(x_0-\beta) } \right) $$
and if we use the fact, that the Schröder-function is just a geometric series and can be expressed in a closed form we get finally:
$$ \operatorname{hgh}(x_h,x_0) = \log_u \left(
 {x_h-\beta \over x_0 - \beta } \cdot{  \sqrt5 + x_0-\beta 
 \over \sqrt5 +  x_h-\beta } \right) $$
Because $u$ is negative we get complex values for fractional $h$ and this must then be handled by selection of the correct branch of the complex logarithm. 
Also it must be handled, that the trajectory of iterations with real heights form a spiral through the complex plane with one round per 2 units in the height and thus the winding-number for heights $h \ge 2 $ must be reflected, say for the example $hgh(2/3,1) = 2$ we need to introduce that correction writing/branching of log
$$ \cdots = { \log \left({ \sigma(x_h - \beta) \over \sigma(x_0-\beta) } \right) + 2 \pi i\over \log u} $$ otherwise we get "the shortest distance" expressed by a complex height of $\small 0.171595 + 0.560130 i$ instead.     
But that are details which I can now work on by myself. Thanks for the hints - sometimes one needs only an external impulse to recover a unremembered concept ... 
