TITLE: Place $0,1,\dots,n-1$ around a circle such that no two numbers are in relative order?
QUESTION [7 upvotes]: For a positive integer $n$, one may arrange the numbers $0,1,\dots,n-1$ at various points on a circle. The "correct" placement for the numbers is to arrange them in increasing order going clockwise around the circle. Two numbers are said to be in in order relative each other if they correct position relative to each other.
Example: In the image below, the numbers 1 and 3 are in the correct order relative to each other (since there is exactly one number between them) but none of the other numbers are in the correct order relative to any other number.

Question: For which values of $n$ can all of the numbers $0,1,\dots,n-1$ be arranged around a circle such that no two numbers are in relative order?
If $n$ is odd, there is a trivial solution. We may place $0$ at the top, then place the remaining numbers in reverse order: $0, n-1, n-2, \dots, 2, 1$. See the example below for $n=7$.

This solution no longer works if $n$ is even, since this would place $n/2$ in correct relative order with $0$.
In fact, I believe there is no possible way to do this if $n$ is even, but I do not have a proof yet. How might one prove this?

REPLY [5 votes]: As mentioned by @Greg Martin's comment, this is equivalent to finding for which $n$ there exists a permutation $[a_1, \ldots, a_n]$ of $[1, \ldots, n]$ such that $[a_1 - 1, \ldots, a_n - n]$ contains only distinct elements modulo $n$ (i.e. is itself a permutation of $[1, \ldots, n]$.
Suppose for contradiction that $n$ is even and this is the case. Then the sum of the elements of $[a_1 - 1, \ldots, a_n - n]$ should be the same as the sum of the elements of $[1, \ldots, n]$, modulo $n$.
$$\sum_{i=1}^n a_i - i \equiv\sum_{i=1}^n i= \frac{n(n+1)}{2}\pmod{n}.$$
The LHS is clearly equal to $0$, but the RHS is nonzero since $\frac{n(n+1)}{2}$ is not divisible by $n$ when $n$ is even (it has one less factor of $2$ in its prime factorization).