Note: $f^n$ denotes the iteration of composition, e.g. $f^3(x)=(f\circ f\circ f)(x)$
I've noticed that particular functions have a certain property where for some number $n$ the iterations of the function cycle through a set of values so that $f^{m+n}(x)=f^m(x)$ for all $m\in\mathbb{N}$. For example, if $f(x)=1-\frac{1}{x}$: $$f^1(x)=1-\frac{1}{x}$$ $$f^2(x)=1-\frac{1}{1-\frac{1}{x}}=\frac{1}{1-x}$$ $$f^3(x)=1-\frac{1}{1-\frac{1}{1-\frac{1}{x}}}=x$$ $$f^4(x)=1-\frac{1}{x}$$ So the cycle has a period of $n=3$. Is there a name for this property, and where can I find more information? Also, are there any cases where the 'period' $n$ varies as a function of the iterate $m$?
Edit:
As others have pointed to in the comments, the property I am describing can be stated succinctly by $F^n(X)=X$* for some $n$, and can apply to functions as well as operators on functions.
Since this extends rather naturally to integer $n$, idempotence and involution would be examples with periods $1$ and $2$, respectively.
If matrix multiplication is used to represent the composition of functions, then the property in question applies to any $M$ such that $M^n=\pm I$ for some $n$. As Will Jagy pointed out, in the example $f(x)=1-\frac{1}{x}$, $M^3=-I$ is given by the Moebius transformation $f(x)=\frac{x-1}{x+0}$.
Given how incredibly general this property is and the number of things to which it applies there is absolutely no way that I am the first person to notice it. There has to be a book or a paper somewhere, right?
*In retrospect, this should have been apparent given that $f^{m+n}=f^m\implies f^n=f^0$
"Corollary"?
If $$\frac{d^nf(x)}{dx^n}=f(x)$$ for some $n\in\mathbb{Z},n\neq0$, then $$\frac{d^{mn}f(x)}{dx^{mn}}=f(x)$$ and $$\int^{m(n-1)}f(x)\ dx^{m(n-1)}=f(x)$$ for all $m\in\mathbb{Z}$
writing your function as the Moebius transformation $$ \frac{x-1}{x+0} $$ we get the matrix $$ M = \left( \begin{array}{rr} 1 & -1 \\ 1 & 0 \end{array} \right) $$ Note that $M^2 -M+I=0$ so $M^3 = -I$ You can do something similar with any $$ M = \left( \begin{array}{rr} a & b \\ c & d \end{array} \right) $$ and $$ \frac{ax+b}{cx+d} $$ such that $M^n = I$ for some $n.$ It is also allowed to have the elements of $M$ complex if you want.