Combinatorial interpretation of the identity $(f \circ f \circ f)(x) = x$ where $f(x) = 1/(1-x)$ for $x\in(-1,1)$

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 $$ T(f(f(f(z)))) = T(z) \implies f^{[3]}= id \, . $$