Let $G$ be a set of connected graphs with the following properties:
i.for all $g \in G$, $g$ has $n+1$ vertices
ii. for every natural numbers from $1$ to $n$, there exists a vertex whose degree is equal to that number.
Prove that $G$ has only 1 isomorphism type. I have never proven anything like this before. Am I supposed to show that given any two graphs in $G$, I can define a bijection between, thus they are under the same equivalence class? If so, how do I approach it
HINT: You have to define not just a bijection, but a bijection that is a graph isomorphism. In other words, if the graphs have vertex sets $V_1$ and $V_2$ and edge sets $E_1$ and $E_2$, respectively, you must show that there’s a bijection $h:V_1\to V_2$ such that for any $u,v\in V_1$, $\{u,v\}\in E_1$ if and only if $\{h(u),h(v)\}\in E_2$.
I recommend that you begin by letting $V_1=\{v_0,v_1,\ldots,v_n\}$, where $\deg v_k=k$ for $k=1,\ldots,n$. Then figure out what the degree of $v_0$ has to be, and which vertices are connected by edges to which other vertices. For instance, $v_n$ must be connected to ... ?