Combinatorial interpretation of the identity $(f \circ f \circ f)(x) = x$ where $f(x) = 1/(1-x)$ for $x\in(-1,1)$ Let $f(x) = 1/(1-x)$. We can interpret this as the generating function of an infinite list of $1$s: $(1, 1, 1, \cdots)$.
Now, let's consider $(f \circ f \circ f)(x)$. We first compute $(f \circ f)(x)$, and use this to compute $f \circ f \circ f$.
\begin{align*}
&(f \circ f)(x) = \frac{1}{1 - \frac{1}{1-x}} \\
&=  \frac{1}{\frac{(1 - x) - 1}{1-x}}
= \frac{1-x}{-x} = \frac{x-1}{x} 
= 1 - \frac{1}{x} \\
\end{align*}
\begin{align*}
&(f \circ f \circ f)(x) = (f \circ f)(f(x)) = 1 - \frac{1}{f(x)} = 1 - (1 - x) = x
\end{align*}
What is going on? It's cool that $f^{\circ 3} = id$, but I don't have an explanation for this. I lose the ability to interpret this as a generating function at step $(2)$: When we compute $f \circ f= 1 - 1/x$, it's unclear to me what this represents.
I have an explanation, but it's not very enlightening.  We can consider $f$ as a mobius transform: $f(z) = (0\cdot z + 1)/(-z+1)$. This gives the matrix $F$ such that $F^3 = I$. So there should be some complex analytic explanation that I am unable to divine.
$$
F = \begin{bmatrix} 0 & 1 \\ -1 &1 \end{bmatrix}; \quad 
F^3 = -I \underset{\small{\text{(projectively)}}}{\simeq} I
$$
I was hoping to view either some Riemann sphere-based explanation (I don't know this very well) or a combinatorial explanation of the above phenomenon.
 A: Here are some interpretations in the context of Möbius transformations, i.e. of the holomorphic automorphism of the extended complex plane (or Riemann sphere). I'll use the notation $f^{[n]}$ for the $n$-th iteration of $f$.
A quick way to see that $f^{[3]} = id$ is the following: The Möbius transformation $f(z) = 1/(1-z)$ maps the points $0, 1, \infty$ to $1, \infty, 0$, respectively, i.e. those three points are permuted cyclically.
 It follows that $f^{[3]}$ fixes these points. Since a Möbius transformation is uniquely determined by the image of three distinct points, $f^{[3]} = id$.
Another way is to use the classification of Möbius transformations. (See also John Olsen: The Geometry of Möbius Transformations for a nice overview). 
$$
 f(z) = \frac{0z + 1}{-1z + 1}
$$
has $\operatorname{tr}^2(f) = (0+1)^2 = 1$, so that it is an elliptic transform. Elliptic transformations are conjugate to a rotation $z \mapsto \lambda z$ with the same (squared) trace:
$$
 \lambda + \frac 1 \lambda + 2 = \operatorname{tr}^2(f) = 1 \\
\implies \lambda^2 + \lambda + 1 = 0 \\
\implies \lambda^3 = 1 \, , \, \lambda  \ne 1 \, .
$$
So $f$ is conjugate to a rotation by a third root of unity, and that implies $f^{[3]} = id$.
That conjugation can also be computed explicitly. The fixed points of $f$ are roots of $z^2-z+1=0$ which are
$$
 z_{1, 2} = \frac 12 (1 \pm i \sqrt 3) \, .
$$ 
It follows that with
$$
 T(z) = \frac{z-z_1}{z-z_2}
$$
the “conjugate” Möbius transformation $g = T \circ f \circ T^{-1}$ has the fixed points $0$ and $\infty$ and therefore is a rotation:
$$
 T(f(z)) = \lambda T(z)
$$
for some constant $\lambda \in \Bbb C$. For $z = \infty$ we get
$$
 -\frac{z_1}{z_2} = T(f(\infty)) = \lambda T(\infty) = \lambda \, ,
$$
i.e. $\lambda = \frac 12 (-1 + i\sqrt 3)$. $\lambda$ is a third root of unity, so that
$$
  T(f(f(f(z)))) =  T(z) \implies f^{[3]}= id \, .
$$
