Vertex-transitive polytope with large facet

Question

Every edge of a triangle contains all but one of the vertices. Every face of a tetrahedron contains all but one of the vertices. Every $(n-1)$-face of an $n$-simplex contains all but one of the vertices.

Every edge of q square contains half of the vertices. Every face of a cube contains half of the vertices. Every $(n-1)$-face of an $n$-cube contains half the vertices.

Is there anything in between the triangle and the square?

Answer

Yes! The dual of the cyclic polytope can be an example if parameters are chosen well.

My knowledge of this combinatorial example is due to Carl Lee; the (poor) exposition is due solely to me.

The polytope is 4-dimensional and its combinatorial automorphism group acts vertex transitively. I'm not sure if the standard embedding as the convex hull on points of the moment curve has full combinatorial automorphism group.

Also, I'm not particularly versed in this area, so I describe it in dual form first:

For every pair of positive integers $n$ and $k$ with $k\geq 2n$ define a polytope $P_{n,k}$ as the $2n$-dimensional (abstract) polytope with vertices the integers $\{1,2,\dots,k\}$ mod $k$ and maximal facets ($(2n-1)$-faces) given by $2n$-sets of the form $$\bigcup_{i=1}^n \{ a_i, a_i +1 \}$$ for integers $a_i$ taken mod $k$ such that result really does have $2n$ elements.

It is not hard to count these, there are $\binom{k-n}{n} + \binom{k-n-1}{n-1}$ of them, and exactly $2\binom{k-n-1}{n-1}$ of them contain the vertex $1$. The cyclic group $\mathbb{Z}/k\mathbb{Z}$ acts vertex transitively on the polytope by acting regularly (by addition) on the vertices. This polytope's full symmetry group is usually the dihedral group of order $2k$ acting naturally on the $k$ points, but is sometimes larger.

The polytope in question is the dual polytope, where $n$ and $k$ are chosen so that the inequality works out.

Specifically, $n=2$ (4-dimensional) and $k=6$ gives vertices $\{1,2,3,4,5,6\}$ and maximal facets $\left\{ \{1,2,3,4\}, \{1,2,4,5\}, \{1,2,5,6\}, \\~~~ \{2,3,4,5\}, \{2,3,5,6\}, \{2,3,6,1\}, \\~~~ \{3,4,5,6\}, \{3,4,6,1\}, \{4,5,6,1\} \right\}$

This polytope has 9 facets, and each vertex is contained in 6 maximal facets.

The dual polytope has 9 vertices, and each maximal facet contains 6 vertices. (Yay!)

The combinatorial automorphism group of the cyclic polytope is a wreath product $$S_3 \wr S_2 = \langle (1,3,5), (1,3), (1,2)(3,4)(5,6) \rangle$$ and one can check explicitly that this acts transitively on the maximal facets. Hence in the dual, the combinatorial automorphism group is vertex transitive.