Vertex Condition for Isomorphic Graphs

Theorem
Let $$G_1$$ and $$G_2$$ be isomorphic graphs.

Then the degrees of the vertices of $$G_1$$ are exactly the same as the degrees of the vertices of $$G_2$$.

Proof
Let $$\phi: V \left({G_1}\right) \to V \left({G_2}\right)$$ be an isomorphism.

Let $$u \in V \left({G_1}\right)$$ be an arbitrary vertex of $$G_1$$ such that $$\phi \left({u}\right) = v \in V \left({G_2}\right)$$.

Let $$\deg_{G_1} \left({u}\right) = n$$.

We need to show that $$\deg_{G_2} \left({v}\right) = n$$.

As $$\deg_{G_1} \left({u}\right) = n$$, there exist $$u_1, u_2, \ldots, u_n \in V \left({G_1}\right)$$ which are adjacent to $$u$$.

Every other vertex of $$G_1$$ is not adjacent to $$u$$.

Let $$\phi \left({u_i}\right) = v_i$$ for $$1, 2, \ldots, n$$.

Because $$\phi$$ is an isomorphism, each of the vertices $$v_1, v_2, \ldots, v_n \in V \left({G_2}\right)$$ are adjacent to $$v$$.

Similarly, every other vertex of $$G_2$$ is not adjacent to $$v$$.

Thus $$\deg_{G_2} \left({v}\right) = n$$.

This applies to all vertices $$u \in V \left({G_1}\right)$$.

Hence the result.

Note
It does not necessarily follow that if the degrees of the vertices of $$G_1$$ are exactly the same as the degrees of the vertices of $$G_2$$, then $$G_1$$ and $$G_2$$ are isomorphic.