Very symmetric convex polytope Let $C_n$ be the convex polytope in ${\mathbb R}^n$ defined by the inequalities
(in $n$ variables $x_1,x_2, \ldots ,x_n$) :
$$
x_i \geq 0, x_i+x_j \leq 1
$$
(for any indices $i<j$). 
Denote by $E_n$ the set of extremal points of $C_n$. We have a natural action
of the symmetric group $G={\mathfrak S}_n$ on $C_n$, and hence on $E_n$ also. So we have
a quotient set $\frac{E_n}{G}$. Here are some questions about it, in decreasing order
of difficulty :
1) Is a simple description of $\frac{E_n}{G}$ known in general ?
2) What is the asymptotic behaviour of the sequence $(|\frac{E_n}{G}|)_{n \geq 2}$ ?
3) Is the sequence $(|\frac{E_n}{G}|)_{n \geq 2}$ bounded ?

Here is a proof that the complete description suggested in Andreas Blass'
answer is indeed correct.
It is clear that all the points listed are indeed extremal. Conversely, let
$v=(x_1,x_2,x_3, \ldots ,x_n)$ be an extremal point in $C_n$. We may assume
wlog that $x_1\geq x_2 \geq x_3 \geq \ldots x_n$.
If $x_1=0$, $v$ is the zero vector : this is case (1).
We can therefore assume $x_1 \gt 0$. Let $r$ be the largest index satisfying
$x_r \gt 0$.
If $r\geq 3$, let $p_r$ be the vector with the $r$ first coordinates are equal
to $\frac{1}{2}$ and all the others equal to $0$.  Then $v$ is a convex combination of the zero vector and $p_r$. Since $v$ is extremal, we deduce $v=p_r$ : this is case (3).
If $r=2$, then $v$ is a convex combination of the first two vectors of the canonical
basis. Since $v$ is extremal, $v$ must be one of those two : this is case (2).
Finally, if  $r=1$, then $v$ is a convex combination of the zero vector and the first  vector of the canonical basis. Since $v$ is extremal, $v$ must be one of those two, and we are in case (1) or (2).
