I don't know if this is known, but I was fiddling around with this equality :
$$f:(-1,1)\to (-1,1)\quad \text{satisfies}\quad(f(z)+1)^s=\sum\_{j=0}^{\infty}\dbinom{s}{j}f(z^k)
\quad \forall z\in (-1,1),s\in \mathbb{R}$$
Also you may assume $f$ is differentiable infinitely many times in $(-1,1)$. 
Now my questions are :


1. What functions work, if we impose, $f$ must be monotone and $f\in \mathcal{C}^{\infty}$?
2. What functions work, if $f$ must be continuous, but not necessarily differentiable?
3. And finally, what functions work when no other condition except $f\in \mathcal{C}^{\infty}$ is given?


Some ideas on this would be helpful. Thanks.



A functional equality
By MassiveJack's comment it follow that
$$f(z^{\frac{p}{q}})=(f(z))^{\frac{p}{q}}$$
for all $p,q \in \mathbb{N}$ and $0<z<1$ and by continuity
$$f(z^\alpha)=(f(z))^\alpha$$
for all $0<z<1,\alpha>0$.
Now put $z=e^{-1}$ and $\alpha=-\log(x)$ then
$$f(x)=f(e^{-1})^{-\log(x)}=x^{-\log(f(e^{-1}))}$$
hence $f$ is of the form $f(x)=x^a$ on $(0,1)$.
As $f(z)^2=f(z^2)$ and $f$ is continuous we have $f(-z)=f(z)$ or $f(-z)=-f(z)$ for all $-1<z<0$. Now it is easy to answer the three questions:


1. $f(z)=x^n$ with $n$ odd.
2. $f(z)=z^\alpha$ for $z>0$ and $f(z)=|z|^\alpha$ or $f(z)=-|z|^\alpha$ for all $z<0$ with $\alpha>=0$.
3. $f(z)=x^n$ with $n \in \mathbb{N}$.


