two fixed points, same fractional iteration Suppose a function f(z) has two fixed points, one repelling, and the other attracting.  Call the repelling fixed point f(-1)=-1, and the attracting fixed point f(+1)=+1.  I'm interested in functions where the fractional iterates are the same, developed from either fixed point.
We can generate fractional iterates, $g_{-1}(z)=f^{oz}$ from the Schroeder function of f(z) developed around the fixed point of -1, and also from the fixed point of +1, $g_{+1}(z)=f^{oz}$.  For what functions "f" will the two fixed points agree on their fractional iterates, such that $g_{-1}(z)=g_{+1}(z+k)$, where "k" is a constant?  
The only case I can find that works is $f(z)=\frac{z+c}{1+cz}$, where $0<|c|<1$, and the inverse function is $f^{-1}(z)=\frac{z-c}{1-cz}$.  Then $g(z)=\tanh(z\tanh^{-1}(c))$, which is derived using the tangent angle sum equation.  Are there any other functions f with symmetrical fractional iterates from both fixed points, or is this function family of functions the only functions with symmetrical fractional iterates from both fixed points?
I know of one other case, iterating z^2, involving a super-attracting fixed point of zero, and a repelling fixed point of 1.
 A: A complete answer to this question can be found in the paper
MR0729400 
S. Dubuc, 
Étude théorique et numérique de la fonction de Karlin-McGregor, 
J. Analyse Math. 42 (1982/83), 15–37.
and references in it. Shortly speaking there are no other rational functions with this property. But of course there are functions on $[-1,1]$, and they are completely described in the references to this paper. 
A: This basically boils down to whether $f'(-1)=\frac{1}{f'(1)}$.
You can take an arbitrary, injective and anlytic inside the open interval (-1,1), function $\psi$ analytic at $-1$ with $\psi(-1)=0$ (and $\psi'(-1)\neq 0$) and $\frac{1}{\psi}$ being analytic at 1 with $\frac{1}{\psi}(1)=0$ and $\left(\frac{1}{\psi}\right)'(1)\neq 0$ (an example of such a function could be tangent (with 0 instead of -1 though)), then the functions $f_t$
$$ f_t(x)=\psi^{-1}(c^t\psi(x)) $$
(you need to choose the correct branch of $\psi^{-1}$ for $x>1$)  are a continuous iteration group, having fixed points at -1 and 1, have derivatives $c^t$ and $c^{-t}$ at the fixed points, and are analytic on the interval (-1,1) including the fixed points.
A discussion of such an example can be found here
This sounds different from the result of Karlin & McGregor who prove that for a certain function class the only such functions can be the linear fractional functions. But the functions described above (if not linear fractional) are not of this class (his class does not allow functions with branch points).
