# Finding the order of a graph’s automorphism group

Given any digraph $$D$$ and an arbitrary integer $$n\in\mathbb{N}$$ if we define for every vertex $$v\in V(D)$$ a set $$S_v=\{(v,1),(v,2),(v,3),\cdots (v,n)\}$$ so that $$\{S_t:t\in V(D)\}$$ is a family of pairwise disjoint, $$n$$ element sets. Then if I define a digraph $$H$$ with $$\small V(H)=\bigcup_{t\in V(D)}S_t$$ and $$\small E(H)=\bigcup_{(u,v)\in V(D)}S_u\times S_v$$, would the order of the automorphism group of $$H$$ be given by $$|\text{Aut}(H)|=|\text{Aut}(D)|(n!)^{|V(D)|}$$?

I’m not sure if I made an error, my reasoning was that every automorphism $$\sigma\in\text{Aut}(D)$$ could be extended to $$n!$$ automorphisms of $$D$$ for each $$v\in V(D)$$ by letting $$\rho_{\pi}((v,k))=(\sigma(v),\pi(k))$$ for each permutation $$\pi\in S_n$$ thus letting $$v$$ vary over the vertices in $$V(D)$$ we see that in each case a different automorphism is formed for every choice of $$\pi\in S_n$$ which generate a total of $$(n!)^{|V(D)|}$$ different automorphisms for each $$\sigma\in\text{Aut}(D)$$, so I concluded $$|\text{Aut}(H)|=|\text{Aut}(D)|(n!)^{|V(D)|}$$

Edit: Just so there is no confusion, for an arbitrary digraph $$G$$ when I write $$\text{Aut}(G)$$ I mean that:

$$\text{Aut}(G)=\small\left\{\sigma\in \text{Sym}(V(G)):\forall x,y\in V(G)\left[(x,y)\in E(G)\iff (\sigma(x),\sigma(y))\in E(G)\right]\right\}$$