Hello, I am referring to the second and third graphs. The second graph should have 60 isomorphisms but I can't see how. I thought it should be (5 choose 1) for a and then (4 choose 2) for the remaining square that gives us a total of 5*6 graphs. Also for the second, the first automorphism is the identity one but what is the second one?
For the third graph, we have (5 choose 3) ways of choosing the triangle and 2 ways of choosing b and d and 2 ways of choosing c and e. This gives us 10*2*2, not 12. Please correct my logic.
Thank you!
For the isomorphisms on the second graph, we start by choosing the first edge, the tail of the graph. There are $\binom{5}{1} = 5$ ways of choosing this edge. We then choose an endpoint of that edge to be incident to the square. There are $\binom{2}{1} = 2$ ways of doing that. Then we choose the square in $\binom{4}{2}$ ways. That gives us $60$.
The two automorphisms for the second graph are the identity and reflection. Move the $\{a, d\}$ edge to $\{a, b\}$ instead.
For the third graph, the number of automorphisms is related to the Dihedral group. We have $5$ automorphisms from rotation $C_{5}$, and then we reflect $C_{5}$ and rotate again to get $5$ more automorphisms. So $Aut(C_{n}) = D_{2n}$, and so we have $2n$ automorphisms.
I don't immediately see the number of graphs for $C_{5}$. If I see it, I'll edit and get back to you.
Edit: So the way I got the cycle is as follows. I chose three vertices for the "triangle" figure at the top, then permuted them to get $60$ ways to form that top figure. I then divided out by the number of rotations for $C_{5}$, which is $5$. This gave me $\frac{60}{5} = 12$. I'm not entirely confident in my answer, so I'd love if someone else could confirm or offer an alternative.