Sets $f_n\in A_f$ where $f_{n+1}=f_n \circ S \circ f^{\circ (-1)}_n$ and operator $\alpha(f_n)=f_{n+1}$ Let's start with a function on the Reals (in this case for $x=0$ is not defined): for example $f(x)=b/x$, $x \in \mathbb R$
I define:

$$f_0:=f$$
$$f_{n+1}:=f_n \circ S \circ f^{\circ (-1)}_n$$

(where $S(x)=x+1$ is the successor function)
Then the I define the set $A_f$:

$$f \in A_f$$
$$f_n\in A_f $$

We can see for $f_0(x)=b/x$ that
$$f_1(x)={ bx \over b+x}$$
I don't know really what I can do, but I started with some observations
$A_f \subset \mathcal P(\mathbb R^2)$
then we can define a "function" $\alpha:A_f \rightarrow A_f$ 
$\alpha (f_n)=f_{n+1}$
and we have that $f_n$ is a superfunction of $f_{n+1}$
$f_n(x+1)=f_{n+1}(f_n(x))$ 
but that do not help me to understand if the procedure ("functions-valued function" $\alpha$) will never end or it give me always new functions... in other words if the set $A_f$ is finite.
I know that for some fucntions it "ends", obvious example is $\alpha(S)=S\circ S \circ S^{\circ -1}=S$ but I don't know more about this "function" $\alpha$. 
Well, my questions are:

1) What I'm talking about?(function over a set of functions) In which field of mathematics I can find
  this problem and a solution?(references?)
2)What are the properties of $f_n(x)$ (depending on the choice of
  $f$)?
3) Is the set $A_f$ finite or infinite or in other words, the procedure $\alpha$ "ends"? (how can I find $|A_f|$
  for other functions $f$?) 
4)If the function is $f_0(x)=b/x$ then there is a way to find general
  formula for $f_n(x)$?

I bet something like the iniectivity and surjectivity of the functions $f$ have great influence in the costruction of $A_f$...
I'm sorry for my bad english and grammar, thanks in advance
UPDATE 15-04
One of my question is if the sequence $f_n$ ends for every $f$, for some function it does I put some example I found and I hope make my question more clear:
$h_0(x):=$$(x-a)\over b$, $h_1(x)=$$bx+1\over b$, $h_2(x)=x+1$ 
$\alpha(h_2)=\alpha(S)=S$ so the set $A_h$ has only three elements because $h_k=h_2$ for $k\gt 2$ 
If I want to prove if every $A_f$ is finite then i must prove that for every function exist a $m\in \mathbb N$ such that $\alpha(f_m)=f_m$ and in other words $|A_f|=m+1$, this is the way to find an answer? 
I'm think about it and the only function I know satisfies $\alpha (g)=g$ is $S$. 

3') Maybe is right say that $A_f$ is finite only if $S\in A_f$?

Another example of $A_g$ finite is $g_0(x)=a^x$
$g_1(x)=ax$, $g_2(x)=a+x$ and $g_3(x)=x+1$
in that case $|A_g|=3+1=4$.
 A: To answer your 4 questions briefly:


*

*Your $\alpha$ is an operator from a space of real valued functions to itself. You probably require the functions to be bijective in some sense (in order for $f^{-1}$ to even make sense), perhaps with both domain and codomain a subset of the reals. 

*By the recursive relation, each $f_n(x)$ is also bijective in the same sense. The behaviour of the sequence $f_n$ seems to depend greatly on the initial choice of $f_0$.

*$A_f$ may or may not be finite, depending on the initial choice of the function $f_0$. I will show an example of $f_0$ which leads to an infinite set $A_f$ below.

*Yes. When $f_0(x)=\frac{b}{x}$, we have $f_1(x)=b+\frac{-b^2}{x+b}, f_2(x)=b-b^2+\frac{-b^4}{x+(-b^2-b)}$ and $$f_n(x)=b-\sum_{i=1}^{n-1}{b^{2^i}}+\frac{-b^{2^n}}{x+(-b^{2^{n-1}}+\sum_{i=1}^{n-2}{b^{2^i}}-b)}, n \geq 3$$
In fact, we shall give a general formula for the case where $f_0(x)=a+\frac{b}{x+c}$. This will be an example which leads to an infinite set $A_f$.
These points shall be further elaborated on below.
We have $\alpha(f(x))=f(f^{-1}(x)+1)$.
Result 1: $S(x)=x+1$ is the only function that maps to itself under $\alpha$.
Proof: First note that $\alpha(S(x))=S(x)$. If $\alpha(f)=f$, then $f(x)=\alpha(f(x))=f(f^{-1}(x)+1)$. Replacing $x$ by $f(x)$ gives $f(f(x))=f(x+1)$. Taking $f^{-1}$ on both sides gives $f(x)=x+1$, so indeed, $S(x)=x+1$ is the only function that maps to itself under $\alpha$.
Result 2: $\alpha(ax+b)=x+a, a \not =0$.
Proof: Consider $f(x)=ax+b$. Then $f^{-1}(x)=\frac{x-b}{a}$, so $\alpha(ax+b)=a(\frac{x-b}{a}+1)+b=x+a$. 
As you noted, this easily leads to $\alpha^2(ax+b)=\alpha(x+a)=x+1$.
Result 3: Suppose $f(x)=ag(x)+b, a \not =0$. Then $\alpha(f(x))=a\alpha(g(x))\circ [\frac{x-b}{a}]+b$.
Proof: We have $g^{-1}(x)=f^{-1}(f(g^{-1}(x)))=f^{-1}(ag(g^{-1}(x))+b)=f^{-1}(ax+b)$. Thus $\alpha(f(x))=f(f^{-1}(x)+1)=f(g^{-1}(\frac{x-a}{b})+1)=ag((g^{-1}(\frac{x-a}{b})+1)+b=a\alpha(g(x))\circ [\frac{x-b}{a}]+b$.
Result 5: $\alpha(a+\frac{b}{x+c})=(a+b)+\frac{-b^2}{x+(b-a)}$.
Proof: Suppose $f(x)=a+\frac{b}{x+c}$. Then $f^{-1}(x)=-c+\frac{b}{x-a}$, so $$\alpha(a+\frac{b}{x+c})=a+\frac{b}{(-c+\frac{b}{x-a}+1)+c}=a+\frac{b(x-a)}{x+(b-a)}=(a+b)+\frac{-b^2}{x+(b-a)}$$
Result 6: Suppose $f_0(x)=a+\frac{b}{x+c}$. Then $f_1(x)=(a+b)+\frac{-b^2}{x+(b-a)}, f_2(x)=a+b-b^2+\frac{-b^4}{x+(-b^2-b-a)}$ and $$f_n(x)=a+b-\sum_{i=1}^{n-1}{b^{2^i}}+\frac{-b^{2^n}}{x+(-b^{2^{n-1}}+\sum_{i=1}^{n-2}{b^{2^i}}-b-a)}, n \geq 3$$
Proof: We proceed by induction on $n \geq 3$, using Result 5. When $n=3$, we have 
\begin{align}
\alpha(f_2(x))=\alpha(a+b-b^2+\frac{-b^4}{x+(-b^2-b-a)}) & =(a+b-b^2)+(-b^4)+\frac{-(-b^4)^2}{x+((-b^4)-(a+b-b^2))} \\
& =a+b-(b^2+b^4)+\frac{-b^8}{x+(-b^4+b^2-b-a)}
\end{align}
so the statement holds for $n=3$.
Suppose that the statement holds for $n=k \geq 3$. Using the induction hypothesis and Result 5, we have:
\begin{align}
f_k(x) & =a+b-\sum_{i=1}^{k-1}{b^{2^i}}+\frac{-b^{2^k}}{x+(-b^{2^{k-1}}+\sum_{i=1}^{k-2}{b^{2^i}}-b-a)} \\
f_{k+1}(x) & =\alpha(f_k(x))=(a+b-\sum_{i=1}^{k-1}{b^{2^i}})+(-b^{2^k})+\frac{-(-b^{2^k})^2}{x+(-b^{2^k}+\sum_{i=1}^{k-1}{b^{2^i}}-b-a)} \\
& =a+b-\sum_{i=1}^{(k+1)-1}{b^{2^i}}+\frac{-b^{2^{k+1}}}{x+(-b^{2^{(k+1)-1}}+\sum_{i=1}^{(k+1)-2}{b^{2^i}}-b-a)}
\end{align}
We are thus done by induction.
It is easy to see that we can choose $a, b$ so that the expression obtained for $f_n(x)$ never repeats, so this implies that $A_f$ is infinite.
More comments: If we were to start with a function such as $f_0(x)=x^3$, the resulting $f_n(x)$ would not be nice. We have for example $f_1(x)=(\sqrt[3]{x}+1)^3, f_2(x)=(\sqrt[3]{(\sqrt[3]{x}-1)^3+1}+1)^3$.
Also, the general problem of determining bijective solutions of $f_k(x)=f_0(x)$ for $k \geq 2$ does not seem to be very tractable. For $k=2$, the conditions are relatively simple, and some work is provided below.
$$g(x)=\alpha(f(x))=f(f^{-1}(x)+1), f(x)=\alpha(g(x))=g(g^{-1}(x)+1)$$
$$f^{-1}(x)=f^{-1}(g(g^{-1}(x)))=f^{-1}(f(f^{-1}(g^{-1}(x))+1))=f^{-1}(g^{-1}(x))+1$$
$$g^{-1}(x)=f(f^{-1}(x)-1)$$
$$f(x)=g(g^{-1}(x)+1)=f(f^{-1}(g^{-1}(x)+1)+1)$$
$$x=f^{-1}(g^{-1}(x)+1)+1$$
$$f(x-1)=g^{-1}(x)+1=1+f(f^{-1}(x)-1)$$
$$f(f(x)-1)=1+f(x-1)$$
Now any solution of $f_2(x)=f_0(x)$ must satisfy $f(f(x)-1)=f(x-1)+1$, and should be bijective as well. We could throw in continuity, if that is necessary. However for larger $k \not =2$, this approach quickly becomes tedious.
