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David E Speyer
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Partial progress: Let $V$ be the vertex set of $P$, let $E$ be the set of directed edges and let $X$ be the set of ordered pairs of distinct elements of $V$. Let $G$ be the group of combinatorial symmetries of the edge graph and let $\Gamma \subset G$ be the group of geometric symmetries of the polytope. So it is assumed that $E$ is a single orbit for both $G$ and $\Gamma$ acting on $X$. I claim there must be some other $G$-orbit on $X$ which splits into more than one $\Gamma$ orbit. In particular, we must have more than one $G$-orbit on $X$, which means that neighborly polytopes won't work.

Without loss of generality, we may assume that $P$ spans $\mathbb{R}^d$ and the centroid of $P$ is at $\vec{0}$, so action of $\Gamma$ extends uniquely to a linear action on $\mathbb{R}^d$.

Proof: Suppose to the contrary that $G$ and $\Gamma$ have the same orbits on $X$. Let the permutation representation $\mathbb{C} V$ decompose into $G$-irreps as $\bigoplus V_i^{\oplus c_i}$, and decompose into $W$-reps as $\bigoplus W_i^{\oplus d_i}$. It is well known that $\sum c_i^2 = \dim \mathrm{Hom}_G(\mathbb{C} V, \mathbb{C} V) = |V^2/G| = 1+|X/G|$. So the hypothesis that $|X/G| = |X/\Gamma|$ means that $\sum c_i^2 = \sum d_i^2$ and, thus, all the $V_i$ stay irreducible as $\Gamma$-reps.

The map taking the basis vector $e_v$ of $\mathbb{C} V$ to the vertex $v$ of the polytope $P$ gives a $\Gamma$-equivariant linear surjection from $\mathbb{C} V$ to $\mathbb{C}^d$. So $\mathbb{C}^d$ can be identified with a $\Gamma$ summand of $\mathbb{C} V$. But the above computation shows that every $\Gamma$ summand is also a $G$-summand, so the $\Gamma$ action extends to a $G$ action, contradiction. $\square$.

On reflection, it would have been better to use $\mathbb{R}V$ instead of $\mathbb{C}V$ and talk about real, complex and quaternionic representations, but I believe it comes out the same in the end.

So we want a graph $(V,E)$ with arc-transitive symmetry group $G$, and a subgroup $\Gamma$ of $G$ which is still arc-transitive but has more orbits on $X$. Such graphs definitely exist. As one example, let $(V,E)$ be the Hamming $n$-cube, whose symmmetry group is $S_n \ltimes C_2^n$ (here $C_2$ is the cyclic group of order $n$.) If $H$ is a transitive but not $k$-transitive subgroup of $S_n$ for some $k$, then $H \ltimes C_2^n$ has more orbits on $X$, but all edges of $(V,E)$ remain a single orbit. But I haven't succeeded yet in embedding an example like this as the edge graph of a polytope.

David E Speyer
  • 156.2k
  • 14
  • 419
  • 763