TITLE: Smallest graph with automorphism group the quaternion $8$-group, $Q_8$
QUESTION [32 upvotes]: Frucht's Theorem states that for any finite group $G$ there is a finite (undirected) graph $\Gamma$ for which the automorphism group $\text{Aut}(\Gamma)$ of $\Gamma$ is isomorphic to $G$, and for many small groups $G$ it is not hard to construct a small graph $\Gamma$ whose automorphism group $\text{Aut}(\Gamma)$ is isomorphic to $G$.
For example, $\Bbb Z_1$, $\Bbb Z_2$, and $S_3$ are the automorphism groups of the complete graphs $K_1, K_2$, $K_3$, respectively, $\Bbb Z_2 \times \Bbb Z_2$ is the automorphism group of
,
$\Bbb Z_3$ is the automorphism group of the $9$-vertex graph 
,
$\Bbb Z_4$ and $\Bbb Z_5$ are respectively the automorphism groups of the $12$- and $15$-vertex analogues of that graph given (informally) by replacing the central triangle with a square or pentagon, $\Bbb Z_6$ is the automorphism group of the $12$-vertex graph
 ,
and $\Bbb Z_7$, $\Bbb Z_8$, and $\Bbb Z_9$ are the respectively the automorphism groups of the $14$-, $16$-, and $18$-vertex analogues of that graph given by replacing the hexagons with heptagons, octagons, and nonagons.
This accounts for all groups of order $< 10$ (in fact, except for the example for $\Bbb Z_4$, these examples are all minimal w.r.t. vertex count, at least among connected graphs) except $\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$, $\Bbb Z_3 \times \Bbb Z_3$, and the quaternion $8$-group, $Q_8$. One can modify without much fuss the above examples to produce graphs with automorphism groups the abelian groups in this list, but it is not so clear (to me) to see how to construct a reasonably small graph with automorphism group $Q_8$:

What is the minimal number of vertices of a graph $\Gamma$ for which $\text{Aut}(\Gamma) \cong Q_8$? What is an example of such a graph that achieves this minimum? Among such graphs, which has (have) the minimal number of edges?

We can given an upper bound for the vertex count: A theorem of Babai says for all groups $G \not\cong \Bbb Z_3, \Bbb Z_4, \Bbb Z_5$ there is a graph $\Gamma$ with $\text{Aut}(\Gamma) \cong G$ such that the action of $\text{Aut}(G)$ has only two orbits, and so in particular has vertex count $V(\Gamma)$ no larger than $2|G|$; thus, there is a graph $\Gamma$ with $\text{Aut}(\Gamma) \cong Q_8$ and $V(\Gamma) \leq 16$. (I expect that $16$ is also the minimum achievable.)
One can attempt to solve this by exploiting a constructive proof of Frucht's Theorem: Roughly, one starts with a Cayley graph of $G$ (which is directed and colored according to generator), which has automorphism group (of directed, colored graphs) $G$. Then, one replaces each edge with a suitable (undirected) graph, choosing a different replacement graph for each generator, to avoid introducing new symmetries.
It is not clear, however, that this approach can produce a graph with $\leq 16$ vertices and automorphism group $Q_8$. The simplest Cayley graph of $Q_8$ is
,
but this already has $16$ edges, and it's hard to see how one could suitably replace them without added more than $8$ new vertices in total.
This answer, by the way, gives a graph with $32$ vertices with automorphism group $Q_8$ (acting with $4$ orbits on the graph):
.
Incidentally, one can use the above approach to improve upon this result, i.e., construct a graph $\Gamma$ with $< 32$ vertices and automorphism group $Q_8$: Let $S(p, q)$ denote the graph with three vertices $a, b, c$, $p$ edges between $a$ and $b$, and $q$ edges between $b$ and $c$. Then, we can replace the red arrows in the above Cayley graph with $S(p, q)$ (appropriately oriented) and the green arrows with $S(p', q')$, where $p, q, p', q'$ are chosen to avoid introducing new automorphisms. Each replacement adds one vertex, so the resulting graph has $24$ vertices. (This graph is, however, not simple, and so is less than entirely satisfying.)

Babai, L., "On the minimum order of graphs with given group." Canad. Math.
  Bull., 17, pp. 467–470, 1974.
Frucht, R., "Herstellung von Graphen mit vorgegebener abstrakter Gruppe." Compositio Mathematica (in German) 6, pp. 239–250, 1939.

REPLY [3 votes]: What is the minimal number of vertices of a graph $\Gamma$ for which $\text{Aut}(\Gamma) \cong Q_8$? What is an example of such a graph that achieves this minimum? Among such graphs, which has (have) the minimal number of edges?

There has been progress on this question since I asked it three years ago: While Jim Belk's answer establishes that a graph $\Gamma = (V, E)$ with automorphism group $Q_8$ must have $|V| \geq 16$ (I don't know the earliest reference for this fact), Graves, Graves, & Lauderdale recently showed (2018) that the graphs with $|V| = 16$ and $\operatorname{Aut}(\Gamma) \cong Q_8$ satisfy the sharp bound $|E| \geq 44$.
Probably the article includes an example of a graph realizing the above minimum, but I cannot find an ungated copy and am away from my institution at the moment.
This REU presentation shows that if we allow more vertices we can find a graph with $\operatorname{Aut}(\Gamma) \cong Q_8$ and $|E| = 40$ and that this is a sharp lower bound for graphs with $|V| \leq 20$.

Graves, C.; Graves, S.J.; Lauderdale, L-K. "Smallest graphs with given generalized quaternion automorphism group." J Graph Theory. 87 (2018), 430-442. https://doi.org/10.1002/jgt.22166
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