Notation: $H $ are partitioned into sub-graphs $ H_1,H_2 \cdots H_x$ . We see them in the adjacency matrix of $H$ given below-
$$H = \begin{bmatrix} H_{(x)} & R_{(x, x-1)} & R_{(x,x-2)} & \dots & \dots & R_{(x,1)} \\ R_{(x,x-1)} & H_{(x-1)} & R_{(x-1,x-2)} & \dots & \dots & R_{(x-1,1)} \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ R_{(x,1)} & R_{(x-1,1)} & R_{(x-2,1)} & \dots & \dots &H_{1} \end{bmatrix}$$
The adjacency matrix of graph $H_k \cup H_e$ is $M_{(k,e)}$. $M_{(k,e)} =\left( \begin{array}{ccc} H_e & R_{k,e} \\ R_{k,e}^{T} & H_k\\ \end{array} \right) $, where, $R_{k,e}$ is the non symmetric sub-matrix of adjacency matrix $H$. Each $H_k$ has constant number of vertices.
Set $\alpha_k$ is the set of automrphism of $M_{(k,k+1)}$, i.e. $\alpha_k \subset Aut(M_{(k,k+1)})$ where $1 \leq k+1 \leq x$.
$\sigma$ is an automorphism of graph $H$, i.e. $H^{\sigma}=H$ where $\sigma= \pi_1 \times \pi_2 \times .... \pi_x$ , i.e. $\sigma$ is a direct product of $\pi_1, \pi_2 , .... \pi_x$.
Question:
It is given that, $\pi_1, \pi_2 , .... \pi_x$ exist as product of permutations/automorphisms in corresponding $\alpha_1 , \alpha_2,...\alpha_x $ .
For example,if $\pi_1 = \sigma_1 \times \tau_1$, then, $\alpha_1$ either has $\pi_1$ or both of $\sigma_1, \tau_1$.
To obtain an automorphism of $H$, we construct direct prodect of all $\alpha_k$.
Is it correct? if not then why?