It is evident true that there are at most 4 points are equal-distance in three dimention Euclidean Space.
What about other metric space? This ideal was first generated by my student Yilin Liu, and some results were obtained by us together.

1. In a discrete metric, $X$ is any non-empty set.
$$\rho(x,y) = \left \lbrace \begin{array}{ll} 1 & x \neq y \\ 0 & x=y \end{array} \right.$$
In this metric space,any subset of $X$ is equal-distance.
2. Suppose that $X$ if we define
$$\rho(A,B) = |x_1 - x_2| + |y_1 - y_2| + |z_1 - z_2|$$,
where $A(x_1,y_1,z_1),B(x_2,y_2,z_2)$.
we can find six points satisfy every two of them have some distance,for instance,$(\pm 1,0,0)\;(0,\pm 1,0)\;(0,0,\pm 1)$.
3. Suppose that $X$ if we define
$$\rho(A,B) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 }+ |z_1 - z_2|$$,
where $A(x_1,y_1,z_1),B(x_2,y_2,z_2)$.
It is hard to find more that 4 point which could feed our demand. Fortunately with a skill construction,we finally get five equal-distance point set as follow
$$(1,0,1); \;(-1,0,-1); \;(-\sqrt{2},0,4-\sqrt{2}); \;(-\frac{3\sqrt{2}}{2},2,\frac{\sqrt{2}}{2}); \;(-\frac{3\sqrt{2}}{2},-2,\frac{\sqrt{2}}{2})$$

What more, it remains to be considered in n-dimention space,or any metric and topology space.