TITLE: Every three of $n$ points is the vertices of an isosceles triangle. What is the max of $n$?
QUESTION [12 upvotes]: Suppose that we have $n\ (\ge 3)$ points in the three dimensional space and that every three of the $n$ points is the vertices of an isosceles triangle. Here, suppose that the vertices of an isosceles triangle do not exist on the same line.

Question : What is the max of $n$?

I have the following conjecture.

Conjecture : The max of $n$ is $7$.

We can see that $n=7$ is possible.
(Example 1)
$$A_i\left(\cos\left(\frac{2\pi}{5}i\right),\sin\left(\frac{2\pi}{5}i\right),0\right)\ (i=1,2,3,4,5),A_6(0,0,0),A_7(0,0,1)$$
where $A_1A_2A_3A_4A_5$ is a ragular pentagon.
(Example 2)
$$A_1(1,0,0),A_2(-1,0,0),A_3(0,\sqrt 3,0),A_4(1,0,2),A_5(-1,0,2),A_6(0,\sqrt 3,2),A_7\left(0,\frac{1}{\sqrt 3},1\right)$$
where $A_1A_2A_3A_4A_5A_6$ is a regular triangular prism.
(Example 3)
$$A_1(1,0,0),A_2(-1,0,0),A_3(0,\sqrt 3,0)$$
$$A_4\left(\frac{1+\sqrt 5}{2},\frac{1-\sqrt 5}{2\sqrt 3},\frac{1+\sqrt 5}{\sqrt 3}\right),A_5\left(-\frac{1+\sqrt 5}{2},\frac{1-\sqrt 5}{2\sqrt 3},\frac{1+\sqrt 5}{\sqrt 3}\right)$$
$$A_6\left(0,\frac{2+\sqrt 5}{\sqrt 3},\frac{1+\sqrt 5}{\sqrt 3}\right),A_7\left(0,\frac{1}{\sqrt 3},\frac{3+\sqrt{5}}{2\sqrt 3}\right)$$
where $A_1A_2A_3A_4A_5A_6$ is a regular triangular frustum.
However, I'm facing difficulty in proving the conjecture. Can anyone help?

REPLY [4 votes]: It seems that, if we relax the restriction that the points must not be colinear (a rather artificial restriction IMO), the maximum is 8; with 9 it has been proven impossible. See references here.
The construction for the 8-points set seems to be as follows: 5 points equidistant over a circle or radius $1$ (say, over the $x,y$ plane, with the center at the origin), plus
an additional point in the origin, plus two additional points on $(0,0,\pm 1)$