Given a small graph, how can you manually calculate the number of automorphisms? I thought of seeing the number of nodes of a particular degree and permuting among them, but aren't there other factors to consider, like the node being part of a cycle, etc? For example, this graph:
Suppose b had two more edges going to two different nodes, then you can't map b to g, even though they have the same degree. So how do you do this?
It is in general not even simple to check if two graphs are isomorphic. In your case, we see that $a,b,m$ are the only degree 1 nodes, hence any automorphism must permute them. Next $c$ is characterized by being the only node with degree 1 neighbours, hence $c$ must be a fixed point (which at the same instant shows that $a,b,m$ may be permuted arbitrarily). You may then note that $d$ can be mapped to any of $d,e,f,g$, but then $e$ must be mapped to the common neighbour of $\phi(d)$ and $\phi(c)=c$. Finally you can map $f$ to any of the remaining neighbours of $c$ and the rest, i.e. the images of $g, h, k$ are then determined. We conclude that the automorphism gorup is isomorphic to $S_3\times D_4$.
