TITLE: Can we realize a graph as the skeleton of a polytope that has the same symmetries?
QUESTION [16 upvotes]: Given a graph $G$, a realization of $G$ as a polytope is a convex polytope $P\subseteq \Bbb R^n$ with $G$ as its 1-skeleton. 
A realization $P\subseteq \Bbb R^n$ is said to realize the symmetries of $G$, if for each graph-automorphism $\phi\in\mathrm{Aut}(G)$ there is an isometry of $\Bbb R^n$ inducing the same automorphism on the edges and vertices of $P$.

Question: Let $G$ be the 1-skeleton of a polytope. Is there a realization of $G$ that realizes all its symmetries?

Note that if we know $G$ to be the skeleton of an $n$-polytope, then the symmetric realization does not have to be of the same dimension. In fact, this is not always possible. The complete graph $K_n,n\ge 6$ is realized as a neighborly 4-polytope, but its only symmetric realization is the simplex in $n-1\ge5$ dimensions. However, this is the only example I know of.
One way to construct a counter-example would be to find an $n$-polytope with a skeleton $G$ that is not more than $n$-connected (e.g. $n$-regular), but where $\mathrm{Aut}(G)$ is no subgroup of the point group $O(n)$. Its symmetric realization must be in dimension $\ge n+1$, but it cannot be because of Balinski's theorem.

REPLY [4 votes]: A counterexample is giving by the edge graph of the dual to the polytope of Bokowski, Ewald and Kleinschmidt, described in Gil Kalai's answer. Let $P$ be the BEK polytope, let $P^{\vee}$ be its dual and let $G$ be the edge graph of $P^{\vee}$. The polytope $P$ is simplicial so $P^{\vee}$ is simple and its edge graph $G$ is $4$-regular. 
The symmetry $\phi$ in BEK is a symmetry of the simplicial complex $\partial P$, so it induces a symmetry of $G$.
By a theorem of Gil Kalai, any $4$-polytope $Q$ with edge graph $G$ is combinatorially isomorphic to $P^{\vee}$. So the dual polytope $Q^{\vee}$ of such a $Q$ would be combinatorially isomorphic to $P$. If $\phi$ were realized by a linear symmetry of $Q$, then this would induce a linear symmetry of $Q^{\vee}$, contradicting the result of  Bokowski, Ewald and Kleinschmidt.
It remains to check that $G$ is not the edge graph of a polytope in some dimension $d \neq 4$. Since $G$ is $d$-regular, we would have to have $d<4$ and $d \leq 2$ is clearly impossible. According to Mathematica's PlanarGraphQ function, $G$ is not a planar graph, so it is not the edge graph of a $3$-polytope.
For those who want to experiment on their own, here are the facets (vertices numbered 0 through 9) of $P$:
{{1,2,3,7}, {1,2,4,8}, {9,2,3,8}, {9,2,0,7}, {1,3,4,7}, {2,3,4,8}, {9,3,0,8}, {2,3,0,7}, {1,4,6,7}, {1,5,6,8}, {9,0,6,8}, {9,5,6,7}, {4,5,6,7}, {1,4,5,8}, {0,5,6,8}, {9,0,5,7}, {1,2,6,7}, {1,2,6,8}, {9,2,6,8}, {9,2,6,7}, {3,4,5,7}, {3,4,5,8}, {3,0,5,8}, {3,0,5,7}, {1,4,5,6}, {1,2,3,4}, {9,0,5,6}, {9,2,3,0}}

and the edges of $G$ (vertices numbered 1 through 28).
{{5, 1}, {6, 2}, {6, 3}, {7, 3}, {8, 1}, {8, 4}, {9, 5}, {11, 7}, {13, 9}, {13, 12}, {14, 2}, {14, 10}, {15, 10}, {15, 11}, {16, 4}, {16, 12}, {17, 1}, {17, 9}, {18, 2}, {18, 10}, {18, 17}, {19, 3}, {19, 11}, {19, 18}, {20, 4}, {20, 12}, {20, 17}, {20, 19}, {21, 5}, {21, 13}, {22, 6}, {22, 14}, {22, 21}, {23, 7}, {23, 15}, {23, 22}, {24, 8}, {24, 16}, {24, 21}, {24, 23}, {25, 9}, {25, 10}, {25, 13}, {25, 14}, {26, 1}, {26, 2}, {26, 5}, {26, 6}, {27, 11}, {27, 12}, {27, 15}, {27, 16}, {28, 3}, {28, 4}, {28, 7}, {28, 8}}

The symmetry $\phi$ switches $1 \leftrightarrow 9$, $7 \leftrightarrow 8$, $0 \leftrightarrow 4$ and fixes $2$, $3$, $5$, $6$. Here is a drawing of $G$: