The equation $f^{\circ k}(x) = \mathrm{Id}$ for $x\in E$ is called the Babbage equation and the general solution is given in the following way [M. Kuczma, *Functional equations in a single variable*]:

Let $1 = n_0 < \cdots < n_r = k$ be the complete set of divisors of $k$ and let
$$ E = \cup_{i=0}^{r}\cup_{j=1}^{n_i}U^i_{j} $$
be an arbitrary decomposition of $E$ into disjoint sets such that for every $i$ the sets $U^i_{1},\ldots,U^i_{i_n}$ have the same cardinality. For $1\leq i\leq r$ and $1\leq j\leq n_i-1$, let $f_{ij}$ be an arbitrary one-to-one map of $U^i_j$ onto $U^i_{j+1}$. Then the general solution is given by:

- $f(x) = x$ for $x\in U^0_1$;
- $f(x) = f_{ij}(x)$ for $x\in U^i_j$, $j=1,\ldots,n_i-1$, $i\ge 1$;
- $f(x) = f_{i1}^{-1}(\cdots(f_{i,n_i-1}^{-1}(x))\cdots)$ for $x\in U_{n_i}^i$, $i\geq 1$.

This solution is also valid when $f$ is a function of more that one variable.

In dimension 1, the solution $f$ is continuous if and only if $f$ is monotone, and in this case either $f(x)=x$ or $f$ is a decreasing function such that $f^{\circ 2} = \mathrm{Id}$.

**Question:** In dimension $n\ge 2$, assuming $f$ is continuous, what are the general solutions of the Babbage equation?

We can of course give some special solutions, such as when $n=2$:

- $f(x,y)=(y,x)$ for even $k$.
- $f(x,y)=(f_1(x),f_2(y))$ such that $(f_1)^{\circ k}(x)=x$ and $(f_2)^{\circ k}(y)=y$.