Combinatorial interpretation of why if $f(x) = 1/(1-x)$, then $(f \circ f \circ f)(x) = x$.

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 really 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.

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