Suppose we have a graph described by the adjacency matrix \begin{align*} \begin{bmatrix} 0 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 & 1 \\ 0 & 1 & 0 &0 & 1 & 0 \end{bmatrix}. \end{align*} I'm trying to find all of the automorphisms of this graph. The identity mapping, clearly, is one. Also, the degree of vertices 1, 3, 4, and 6 is $2$, while the degrees of $2$ and $5$ is three. So, we can only map $2$ to itself or to $5$ and can only map $5$ to itself or to $2$. But, if we do that, to maintain the structure of the graph, we also need to swap $1$ and $4$. And, we can also make these two switches and swap $3$ and $6$, which have the same adjacencies, or simply switch $3$ and $6$ and change nothing else. I believe these are the only automorphisms of the graph, as any additional move requires altering connectedness in some way, e.g., swapping either $2$ or $5$, which essentially locks in the remaining moves.
So, I found that the automorphisms are:
\begin{align*} & (1)(2)(3)(4)(5)(6) \\ & (25)(14) \\ & (36) \\ & (25)(14)(36) \end{align*} If I work out the group table for this, I seem to be able to verify both closure under composition and inverses, so it seems that this qualifies as a group. But I can't help but feel that I missed something.
How does this look?
A good way to check your automorphisms is to draw the graph!
I have translated the adjacency matrix into a drawing by labelling the rows/columns as vertex 0, 1, 2, etc. Here we see that we have two choices for placing 1 and 4, which then determines where 0 and 3 are. After that, we have two choices for placing 2 and 5. So the graph has four automorphisms, and your list of them is correct.