When can the Cayley graph of the symmetries of an object have those symmetries?

Question

Let $P$ be an object in $\mathbb{R}^n$ with symmetry group $G$. Let $C$ be the a Cayley graph of $G$.

When can $C$ be embedded in $\mathbb{R}^m$ so that the embedded graph has the same symmetry group $G$ as the original object $P$?

For example, the standard Cayley graph for $D_4$, the symmetry group of the square, can be embedded in the plane with the same symmetry:

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          ^{(Image from J.Gross.)}

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Let us view the Cayley graph as a directed, unlabeled graph.

I would be interested in both positive and negative examples: $P$ such that $C$ can share the same symmetries, and $P$ such that none of its $C$ can realize the same symmetries.

Answer 1

(Answer rewritten in view of better understanding and interesting discussion in comments:)

I first want to show that for "most" finite groups, all Cayley graphs of the group can be embedded into some $\mathbb{R}^n$ such that the symmetry group of the embedded Cayley graph is just $G$ (whether the Cayley graph is directed and labeled or not), and such that the natural action of $G$ on the embedded Cayley graph and the "natural" action on the Cayley graph coincide. (The last point is not true for the example in the question: $D_4$ has two orbits on the vertices of the embedded Cayley graph.)

Babai has shown the following: If the finite group $G$ is neither abelian of exponent $>2$ nor generalized dicyclic, then there is some $v$ in some $\mathbb{R}^n$ with a representation $G\to \mathbf{O}(\mathbb{R}^n)$, such that $|G|=|Gv|$ and such that the symmetry group of the point set $Gv$ is just $G$. (Except for elementary abelian $2$-groups, Babai uses the regular representation $G\to \mathbf{O}(\mathbb{R}^{G})$ in his proof.) If we add line segments between $gv$ and $gsv$ corresponding to elements $s\in S$ of some generating set $S$ of $G$, then the resulting embedded Cayley graph has also symmetry group $G$, whether we consider the arcs as directed and labeled or not. (At least when we use the regular representation, these line segments will not intersect.)
This argument shows that the question concerning geometric embeddings is in fact easier than the question of graphical regular representations.

If, on the other hand, $G$ is abelian of exponent $>2$ or generalized dicyclic, then $G$ has an automorphism sending each group element to its inverse or itself, and which yields a symmetry of all such point orbits $Gv\subseteq \mathbb{R}^n$. This is also a symmetry of the Cayley graphs obtained in the above way, if we forget directions of arcs. On the other hand, in these groups, each generating set must contain elements of order $>2$, so it is not a symmetry of the directed and labeled Cayley graph. But this does not completely clarify the situation for these groups, since there may be other additional symmetries (depending on the representation).

The original version of this answer contained the following special case: Suppose that the object $P\subseteq \mathbb{R}^n$ is a convex polytope such that its symmetry group $G$ acts transitively on the vertices of $P$ (and thus $P$ is the convex hull of an orbit of a point, called an orbit polytope). Then there is some point $v\in \mathbb{R}^n$ such that $|Gv|=|G|$ (only the identity of $G$ fixes $v$) and such that the symmetry group of $Gv$ is only $G$. (See Lemma 1 in Babai's paper or Corollary 5.4 in our paper. In fact, "almost all" points have these properties.) Let $Q$ be the convex hull of $Gv$, which is another orbit polytope. The $1$-skeleton of this polytope is in a natural way (an undirected version of) a Cayley graph of $G$, where the generating set $S\subseteq G$ is the set of elements $s\in G$ such that $v$ and $sv$ are connected by an edge (a $1$-dim. face) of $Q$.
(Also revelant in this context is a paper of Ellis, Harris and Sköldberg (MR2270569 (2008g:20117)).)

Answer 2

[Wiki, as already in the comments]

For the easy negatives, consider the Cayley graph with respect to the generating set $S=G$. The graph obtained is then a complete graph. It can be embedded in $\mathbb{R}^{G} = \{ (a_g)_{g \in G} \mid a_g \in \mathbb{R} \}$ as follows: $g \mapsto \delta_{g\cdot}$ where $\delta_{gh} = 1$ if $h=g$ and $0$ else. The edges between two vertices all have length $\sqrt{2}$.

The permutation matrices are linear isometris of $\mathbb{R}^G$ and will permute the vertices as desired. So the symmetries of the Cayley graph contain $S_{|G|}$ which is strictly larger than $G$ as soon as $|G| \geq 2$.

More generally, as Nick Gill pointed out, this embedding shows that (without bound on $m$, or in fact, as soon as $m \geq |G|$) there is no difference between the "abstract" automorphisms and the "geometric" automorphisms. Positive answers for "abstract" are found in this answer of C. Godsil (again, pointed out by N. Gill) https://math.stackexchange.com/questions/1098115/when-is-the-automorphism-group-of-the-cayley-graph-of-g-just-g

Answer 3

Take $P$ to be a regular polygon on $n$ vertices, with symmetry group the dihedral group $D_n$ of order $2n$. Take the generating set to be all of $D_n$ so that the Cayley graph is the complete graph on $2n$ vertices (the generating set can be without the identity to avoid loops). This complete graph has a nice symmetric embedding in the plane that looks like a regular $2n$-gon with a bunch of lines connecting the vertices of the $2n$-gon through its interior. There will be lots of regular $n$-gons (by taking every other vertex of the $2n$-gon). Alter each edge of one of these interior $n$-gons in a symmetric way (say by adding a bump in the middle of each edge) so that the only plane symmetries of the embedded graph that remain are the symmetries which map the $n$-gon to itself. So this new embedding has plane symmetry group $D_n$, just as $P$ did.