On cubic Cayley graphs of finite simple groups

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Problem: Reconstructed conjecture: every finite group $G$ is a connected $3$-CI-group; i.e. every connected Cayley graph $\mathrm{Cay}(G,S)$ with $S=S^{-1}$, $1\notin S$, $|S|\le 3$, and $\langle S\rangle=G$, is a CI-graph: whenever $\mathrm{Cay}(G,S)\cong \mathrm{Cay}(G,T)$, there is $\alpha\in\mathrm{Aut}(G)$ with $S^\alpha=T$.

Result: The conjecture is false.

Let

$$G=D_8=\langle r,s\mid r^4=s^2=1,\ srs=r^{-1}\rangle$$

be the dihedral group of order $8$. Put

$$S=\{r,r^{-1},s\},\qquad T=\{s,rs,r^2s\}.$$

Both are inverse-closed generating sets of size $3$, so the corresponding Cayley graphs are connected cubic Cayley graphs on $G$.

The graph $\mathrm{Cay}(G,S)$ is the prism $C_4\square K_2$, hence the cube $Q_3$: the elements $1,r,r^2,r^3$ form one $4$-cycle, the reflections form another, and multiplication by $s$ gives a perfect matching.

The graph $\mathrm{Cay}(G,T)$ is bipartite with parts $\langle r\rangle$ and $\langle r\rangle s$. Each rotation is adjacent to exactly three of the four reflections, so it is $K_{4,4}$ minus a perfect matching, also isomorphic to $Q_3$. Hence

$$\mathrm{Cay}(G,S)\cong \mathrm{Cay}(G,T).$$

However no automorphism of $G$ maps $S$ to $T$, because automorphisms preserve element orders: $S$ contains two elements of order $4$, namely $r,r^{-1}$, while every element of $T$ is an involution. Thus $S$ is not a CI-subset.

Therefore $D_8$ is not a connected $3$-CI-group.

Citation: No external result is needed; the counterexample is elementary.