Are these two tournaments on four vertices isomorphic?

Question

I understand that showing an invariant will demonstrate that the two graphs are not isomorphic. I can't find one, but I can't find an isomorphism either. Can you help me settle this?

Answer 1

Here is your isomorphism: $(u_1,u_2,u_3,u_4)\to(v_3,v_1,v_4,v_2)$

Here is how you find it: there are two vertices with outgoing degree $2$ and two with outgoing degree $1$ on both graphs. The ones with outgoing degrees ($u_1$ and $u_2$) have to be mapped to similar ones ($v_1$ and $v_3$). But note that $u_2$ points to $u_1$ and $v_1$ points to $v_3$, so you have to map $u_2$ to $v_1$ and $u_1$ to $v_3$. Apply the same argument to the other two vertices and you will find the whole map.

Answer 2

How many tournaments are there?

The top-ranked answer there suggests persuasively that all tournaments on four vertices with a (directed) Hamilton cycle are isomorphic. Given that both of yours do $(1\to3\to4\to2\to1)$, they are isomorphic.

Once you have that context, it is easier to find the isomorphism. The tournament without its Hamilton cycle is just two arcs. So you can quickly note that $u_3$ is the only vertex with an outdegree of one that leads to the other vertex with an outdegree of one. In the second graph, the vertex with that condition is $v_4$. Following the two Hamilton cycles around, the isomorphism is $$u_1\mapsto v_3, u_2\mapsto v_1,u_3\mapsto v_4, u_4\mapsto v_2$$