They meet the requirements of both having an $=$ number of vertices ($7$).
They both have the same number of edges ($9$).
They both have $3$ vertices of degree $2$ and $4$ of degree $3$.
However, graph two has $2$ simple circuits of length $3$ whereas graph one has only $1$ of length $3$.
Is this not a valid method for checking isomorphism?
My guide says that these two figures are isomorphic.
Best Answer
Once you know, as pointed out in this answer, that $$f(A)=7,\: f(B)=4,\: f(C)=3,\: f(D)=6,\: f(E)=5,\: f(F)=2,\: f(G)=1$$ isomorphism, you can create an animation illustrating how to morph one graph into the other. Let's say that ${vc}_1$ is a list of vertex coordinates for one and ${vc}_2$ is the corresponding list of vertex coordinates for the other. (It's important that the order of the vertex coordinates be dictated by the isomorphism.) We can then morph from one graph to the other using a function like
$$p(t) = t \, {vc}_2 + (1-t) {vc}_1.$$
Here's the result:
Since several folks asked for code generating the animation in the comments, you can find it here: