TITLE: two fixed points, same fractional iteration
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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.