# How to find the automorphism group of the graph mentioned in the question?

Hi, this is my 2nd question in this site. If it is not up to the mark, please do let me know, I will surely try to improve.

The problem is from Graph Theory, specially Automorphism Group. I know that the Automorphism Group of the complement of the complete graph $$\overline{K}_n$$ is isomorphic to $$S_n$$, the symmetric group on $$n$$ symbols.

Also the Automorphism Group of the path graph $$P_4$$ will be $$\{I,(1,4)(2,3\}$$ which isomorphic to $$S_2$$.

My problem is a bit tricky where I am stuck:

Problem: Let $$G$$ be a graph such that $$G$$ is made of $$4$$ components $$G_1,G_2,G_3,G_4$$ where $$G_1=\overline{K}_2,G_2=\overline{K}_3,G_3=\overline{K}_4,G_4=\overline{K}_5$$, i.e. $$G$$ can be obtained from $$P_4$$ in the following way:

The vertex 1 in $$P_4$$ is replaced by $$G_1=\overline{K}_2$$, the vertex 2 in $$P_4$$ is replaced by $$G_2=\overline{K}_3$$, the vertex 3 in $$P_4$$ is replaced by $$G_3=\overline{K}_4$$, the vertex 4 in $$P_4$$ is replaced by $$G_4=\overline{K}_5$$. All the vertices of $$\overline{K}_2$$ are adjacent to all the vertices of $$\overline{K}_3$$, all the vertices of $$\overline{K}_3$$ are adjacent to all the vertices of $$\overline{K}_4$$, and all the vertices of $$\overline{K}_4$$ are adjacent to all the vertices of $$\overline{K}_5$$.

Find the automorphism group of $$G$$ and its order.

My try: I understand that the automorphism group of $$G_1$$ is isomorphic to $$S_2$$,
the automorphism group of $$G_2$$ is isomorphic to $$S_3$$, the automorphism group of $$G_3$$ is isomorphic to $$S_4$$, and the automorphism group of $$G_4$$ is isomorphic to $$S_5$$. Also the automorphism group of $$P_4$$ is isomorphic to $$S_2$$. I also understand that the blocks $$G_1,G_2,G_3,G_4$$ follow the adjacency criterion as in $$P_4$$.

But I am stuck on how to find the automorphism group of $$G$$ from here. I require some help from you all. I am looking forward to some responses from all the learned experts available in MSE.

The vertices in $$G_1,G_2,G_3,G_4$$ are characterized by having degrees $$3,6,8,4$$, respectively. Hence any automorphism must permute vertices only within each $$G_i$$. This makes the automorphism group a subgroup of $$S_2\times S_3\times S_4\times S_5$$. On the other hand, any permutation of $$G_1$$ alone (or $$G_2$$ alone or …) turns out to be an automorphism, hence the group is indeed $$S_2\times S_3\times S_4\times S_5$$.