What arangement of $n$ points in the plane minimizes the dispersion of the distances between them? If $n=3$ the solution is an equilateral triangle where all sides have the same length. In general, there are $N = \frac{n}2(n-1)$ distances between the points. Use the empirical coefficient of variation as a relative measure of dispersion of the distances $x_i$
$$v = \frac1{\bar{x}} \sqrt{\frac1N \sum_{i=1}^N (x_i - \bar{x})^2} \qquad \bar{x} = \frac1N \sum_{i=1}^N x_i$$
Below are some arangements of four points and their values of $v$ (distances with the same length are coloured). I have not found an arangement for $n=4$ with a smaller variation than the square. It is possible to show that for two points at $(\pm 1,0)$ and the two other ones at $(0,\pm y)$ the minimum of $v$ occurs at $y = 1$.
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For $n < 8$ the regular polygons seem to minimize the variation $v = P(n)$. After that it is better to place one point at the center and arange the others evenly around it like in the arangement on the right. In that case, call $v=Q(n)$
$$P(n) = \tan{\left( \frac\pi{2n} \right)} \sqrt{N - \cot^2{\left( \frac\pi{2n} \right)}} \qquad
Q(n) = \frac{\sqrt{\frac{n^2}2 - \left( \cot{\left( \frac\pi{2(n-1)} \right)} + 1 \right)^2}}{\cot{\left( \frac\pi{2(n-1)} \right)} + 1}$$
$$\begin{array}{c|cccc|ccc} n & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline P(n) & 0.172 & 0.236 & 0.277 & 0.3066 & 0.328 & 0.345 & 0.359 \\ Q(n) & 0.268 & 0.269 & 0.287 & 0.3068 & 0.324 & 0.339 & 0.351 \end{array}$$
Are there better strategies for $n \to \infty$? And is there a way to prove an arangement is optimal for a given $n$?
 A: Not quite an answer  (I'd be suprised if this problem can be solved analytically) but, I've playing a little:
This fiddle lets you play for different values of $n$. The minimization algorithm (not optimized!) has a "temperature" parameter (kind of a simulated annealing), higher values lets you escape from local minima.
Empirically, my findings agrees with yours: for $n = 4 \cdots 7$ the regular polygons win. For $n=8, 9, 10, 11$ it's the polygon with a central point - but for $n=11$ the configuration with two internal points is a quite deep (but still suboptimal) local minimum.
For $n=12$, the configuration $(10,2)$ (two internal points) is the optimal one, but $(11,1)$ and $(9,3)$ are relevant competitors.



For larger $n$ things get more complicated, with many similar local mimima.
For example, for $n=21$, the first configuration here $(15,5,1)$ is the main attractor, but there are other three importal local minima, and the last one $(16,5)$ seems to be the optimal.




For even larger values, the points tend to distribute all over the circle, as expected, but with higher concentration over the circumference.
For $n=300$ I get $v=0.4604068$. Not too far from the upper bound limit $0.4834258476$ (where all the points lie on the circumference).

Visually, this appears to support Anders Kaseorg's answer, points seem to correspond to the projection of a uniform distribution over a sphere.
A: If we assume that, in the limit as $n → ∞$, the optimal distribution of points converges to some rotationally symmetric distribution, and optimize a numerical approximation of this distribution, we find outrageously strong numerical evidence that the distribution appears to be the one you get by orthogonal projection onto a plane from the uniform distribution on the surface of a sphere.

(Plot of the predicted vs. optimized inverse CDF of the distance from the origin, normalized such that $\overline x = 1$.)
Consider the segment between two uniformly random points on the sphere of radius $r$ (before projection). Let $a$ be the cosine of the angle between the segment and the projection direction, and let $b$ be the cosine of the spherical angle between the points. Then $a$ and $b$ are independent uniformly random values between $-1$ and $1$. We can compute the planar distance between the projected points as $r\sqrt{1 - a^2}\sqrt{2 - 2b}$. So the mean planar distance is
$$\overline x = \frac14 \int_{-1}^1 \int_{-1}^1 r\sqrt{1 - a^2}\sqrt{2 - 2b}\,da\,db = \frac{πr}{3},$$
and the coefficient of variation is
\begin{multline*}
v = \frac{3}{πr} \sqrt{\frac14 \int_{-1}^1 \int_{-1}^1 \left(r\sqrt{1 - a^2}\sqrt{2 - 2b} - \frac {πr}3\right)^2\,da\,db} \\
= \frac{\sqrt{12 - π^2}}{π} ≈ 0.464601123231588.
\end{multline*}
