Iterative root

Remarks

Example: on $X = \R$, the map $g(x) = x + 1$ is an iterative root of order $3$ of $f(x) = x + 3$. Check: $g^2(x) = x + 2$ and $g^3(x) = x + 3 = f(x)$.

Non-example: $g(x) = 3x$ is not an iterative root of order $3$ of $f(x) = x + 3$, since $g^3(x) = 27x \ne x + 3$.

The roots of order $2$ are the functional square roots. Roots are rarely unique: both $2x$ and $-2x$ are iterative roots of order $2$ of $4x$. Some maps have no continuous root of a given order. If $g \colon \R \to \R$ were continuous with $g^2(x) = -x$, then $g$ would be injective, hence strictly monotone. Then $g^2$ would be increasing, but $x \mapsto -x$ is decreasing. For odd $n$ the same map has the root $g(x) = -x$, since $(-1)^n = -1$.