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\}$