I have $n$ distinct points $x_0,x_1,\ldots,x_{n-1}$ on a circle, with the property that the angle $\theta_{i,(i+1)\%n}$ between consecutive points is the same for all $i\in\{1,\ldots,n\}$. (This angle is measured counterclockwise from $x_i$ to $x_{(i+1)\%n}$.) This does not necessarily imply that two consecutive points $x_i$ and $x_{(i+1)\%n}$ are also spatially consecutive on the circle. For example, we could have $x_0=(1,0)$, $x_1=(0,-1)$, $x_2=(-1,0)$ and $x_3=(0,1)$, with $\theta_{i,(i+1)\%n}=3\pi/2$. Show that
$$ \min_{j\neq i} \theta_{ij} = \frac{2\pi}{n}\quad \forall i\in\{1,\ldots,n\}. $$
Alternatively, show with a counterexample that the statement above does not hold.
Hints and progress so far
Since the angles $\min_{j\neq i} \theta_{ij}$ must add up to the full circle, we know that
$$ \sum_{i=1}^N \min_{j\neq i} \theta_{ij} = 2\pi. $$
Since the angles $\theta_{i,(i+1)\%N}$ must add up to an integer number of circulations, we know that
$$ \sum_{i=1}^N \theta_{i,(i+1)\%n} = 2k\pi \text{ for some $k\in\mathbb{N}$.} $$
Let $x_{\phi(i)}$ be the next marked point on the circle clockwise (not necessarily $x_{i+1}$). We have
$$x_i, x_{\phi(i)}, \ldots, x_{i+1}, x_{\phi(i+1)}, x_{\phi(i)+1},$$
where some points may coincide, but necessarily $x_{\phi(i+1)}$ is before $x_{\phi(i)+1}$. But
$$|x_{i+1}-x_i| = |x_{\phi(i)+1}-x_{\phi(i)}|,$$
thus
$$|x_{\phi(i)+1}-x_{i+1}| = |x_{\phi(i)}-x_{i}|,$$
and
$$|x_{\phi(i+1)}-x_{i+1}| \leqslant |x_{\phi(i)}-x_{i}|.$$
Vice versa inequality is also true, thus $\min_{\theta_{ij}}=|x_{\phi(i)}-x_{i}|$ is constant.