Consider the automorphism group $\text{Aut}(H)$ of a graph $H$. I say that $\text{Aut}(H)$ is disjointly generated if there exists a set of generators $A$ of $\text{Aut}(H)$ such that for each vertex $v$ of $H$, there exists at most one generator $g \in A$ with $g(v) \neq v$.
I'm interested in the size $n_{\text{min}}$ of the smallest graph $H$ such that $\text{Aut}(H)$ is not disjointly generated. Are there any lower bounds (either by mathematical proof or reference to literature) on $n_{\text{min}}$? I'm particularly interested in whether $n_{\text{min}} > 8$.
It is easy to see that $n_{\text{min}} > 4$ by brute-forcing all graphs with $\leq 4$ vertices. However, this approach quickly becomes infeasible to do by hand. One can of course design an algorithm to compute this automatically, but any approach I've come up with is both hard to implement and becomes infeasible even to a very fast computer before $n=8$. I would also accept an answer that discusses a reasonably quick algorithm to verify that $n_{\text{min}} > 8$, but prefer mathematical proofs or references to literature.
If $\operatorname{Aut}(H)$ is disjointly generated, it should be abelian. (Any two generators commute, because any two generators permute disjoint sets of vertices.) So the smallest graph with a non-disjointly-generated automorphism group is $K_3$, because $\operatorname{Aut}(K_3) = S_3$, and $n_{\min} =3$.