What can be said about a function with rotational symmetry of order other than 2? It is well known that an odd function is a function whose graph has rotational symmetry of order $2$ (about the origin).
Suppose the graph of $f:U \to \Bbb{R}$ has rotational symmetry of some higher order $n$, where $U \subset \Bbb{R}$. What can be said about $f$? 
Some things include:


*

*The graph of $f$ must be disconnected (by applying the symmetry of the graph to the vertical line test).

*In particular, if $U$ is an interval, $f$ cannot be continuous.

*Such $f$ exist for certain disconnected $U$, and can be as nice as you like given their domain (e.g., piecewise linear). For a trivial example, let the graph of $f$ consist of the vertices of some appropriately rotated $n$-gon. This can easily be extended to a function on some union of intervals.

*When $n=3$, it is possible to define $f$ on an interval with some deleted points (start with $f(x)=1$ for $x \in (-a,a)$ and extend by symmetry, with $a$ being as large as possible without collision), but at least in this case there's no way to consistently fill in the deleted points.


Is there any such function which is defined on an interval? If so, how badly behaved does it have to be (e.g., can it be piecewise continuous? must it be non-measurable?)
 A: Below are the examples of graphs with 4-fold symmetry. The functions are piecewise continuous (and even piecewise linear, though this could easily have been done otherwise) and defined on entire $\mathbb R$.
The left function is defined on $\mathbb R^+$ as 
$$y=\begin{cases}
0&\text{ if  }x=0\\
1+x&\text{ if  }\lfloor x\rfloor\text{ is even,}\\
1-x&\text{ if }\lfloor x\rfloor\text{ is odd}
\end{cases}
$$
and then continued to negative $x$ in the odd fashion.
The right one is
$$y=\begin{cases}
0&\text{ if  }x=0\\
a\cdot x&\text{ if  }\lfloor\log_ax\rfloor\text{ is odd,}\\
-x/a&\text{ if }\lfloor\log_ax\rfloor\text{ is even}
\end{cases}
$$
where $a>1$ is some parameter.
As you may see, the first function doesn't have any degrees of freedom (except scaling, which is not interesting), but the other one does (the mentioned $a$). One may wonder whether it is possible to accommodate for the six-fold thing by sacrificing that degree of freedom. Now, lo and behold...

The slopes of the linear segments are (top to bottom) $\sqrt5+2\over\sqrt3$,  $\sqrt5-2\over\sqrt3$, and $-{\sqrt5\over\sqrt3}$. The points of discontinuity form a geometric progression, whose common ratio (as luck would have it) is the golden ratio.
So it goes.
