Unless I misunderstood your definition of strong equidistribution I think that your claim that 20 is the maximum cardinality is wrong.
Edit: A trivial example is that $n$ equally spaced points along the equator (with their longitudes differing by $2\pi/n$) are strongly equidistributed according to this definition. You can choose $n$ to be as large as you wish.
For a more interesting example consider the Icosahedral group $G$ of order 120 acting faithfully on $S^2$. If you pick any point $x$ on the surface of the sphere and consider the orbit $C=Gx$, it will have the strong equidistribution. This is because $G\le O_3(\mathbb{R})$ and thus you can turn any point of the orbit $C$ to another by rotating and reflecting the sphere (both of these operations preserve all distances, so strong equidistribution follows).
The same principle can be extended to other dimensions by using a finite subgroup of $O_n(\mathbb{R})$.
Related problems are studied under the umbrella of telcomm applications. I would suspect that a lot of examples of sets with strong equidistribution are to be found within the class of spherical codes. Check out the link to N. J. A. Sloane's database of good spherical codes.
Group codes (orbits of finite groups of orthogonal transformations) are an important subclass.
A few formalities: Let $G\le O_n(\mathbb{R})$ be a finite group, $x\in\mathbb{R}^n$ an arbitrary initial point, then the associated group code is
$$
C=Gx=\{gx\mid g\in G\}.
$$
Because $G$ is a group we have $Gy=C$ for all $y\in C$. Because $G\le O_n(\mathbb{R})$ we have
$$
d(x,y)=d(gx,gy)
$$
for all the vectors $x,y\in\mathbb{R}^n$ and all the elements $g\in G$.
Claim 1. Let $y\in C$ be arbitrary. The distances between closest neighbor $z\in C$ to $y$ is the same for all $y$.
Proof. If $y'\in C$ is the closest neighbor of $y$, and $g\in G$ is such that $gy\in z$ (such an element exists by construction of $C$), then
$$
d(z,gy')=d(gy,gy')=d(y,y'),
$$
so $z'=gy'$ is at the same distance from $z$ as $y'$ from $y$. There cannot be points of $C$ any closer to $z$, because then we could switch the roles of $z$ and $y$ and contradict ourselves.
Claim 2. Let $x_1,x_2,\ldots,x_n$ be all the points of $C$ at minimum distance $d$ from $x$. Let $y\in C$ be arbitrary. Then there are exactly $n$ points of $C$ at distance $d$ from $y$ also.
Proof. Again we select an orthogonal transformation $g\in G$ such that $gx=y$. Then all the points $y_i=gx_i,i=1,2,\ldots,n$ are at distance $d$ from $y$. They are also all distinct. There cannot be more than $n$ points at distance $d$ from $y$, because then we, again, contradict ourselves by reversing the roles of $x$ and $y$.
