Determine isomorphic graphs in a set I am having trouble understanding the following question. Which of the following graphs are isomorphic? 

The answer provided was the graphs $i$ and $iii$ are isomorphic and there are no other isomorphisms. 
This is my answer
Graphs $i,ii,iii$ are all isomorphic as they all have equal number of vertices and for each vertex there is a corresponding vertex (unique) in another graph with the same degree.
Graph $i$
$|V| = 6 \\deg(V)=3$

Graph $ii$
$|V| = 6 \\deg(V)=3$

Graph $iii$
$|V| = 6 \\deg(V)=3$
What am I doing wrong? It feels like I'm missing something simple but I still cant see it. Thanks in advance!
 A: The graph $ii$ doesn't have a cycle of length 3 so it can't be isomorphic to any of the other two graphs. The conditions you listed are necessary but not sufficient for two graphs to be isomorphic.
A: Isomorphism is means of telling if two graphs are the same in some sense. A good way to think of an isomorphic "sameness" is as a "relabeling" or a permutation of the vertices. This is equivalent to the formal definition of a an isomorphism between graphs: 
Definition: Isomorphism of graphs. An isomorphism between graph $G_1(\boldsymbol {V_1},E)$ and $G_2(\boldsymbol {V_2}, E')$ is a bijective (meaning it is both one-to-one and onto) function $f$ from $V_1 \rightarrow V_2$, where if $v_\alpha,v_\beta \in V_1$ are adjacent then $f(v_\alpha),f(v_\beta )\in V_2$ are also adjacent. 
Once can think of this as being a relabeling of the vertex set, because if any two vertexes are adjacent, then if they are relabeled that will not change, and relabeling is obviously bijective. Here is a simple example:

If I move (map) each vertex as I showed (with $F$ mapping to itself), then this would be equivalent to simply moving the labels in a similar manner. However, not all graphs have the nice symmetry of this one, and so we can ask the harder question: is isomorphic to:

The simple response here is to look for an explicit isomorphism (Hint: $F \mapsto L$), but a more enlightening one is to look at how we might move the vertices of the second graph to resemble the first one, so that we might see if they are indeed the same graph just "relabeled". Indeed we simply move one vertex and see they are very similar, and that a natural isomorphism can be drawn:

What underpins this notions of relabeling is put more explicitly in terms of adjacency matrices. Namely, what we are doing is saying: if we permute the adjacency matrix of one graph (meaning we re-order the vertices and their cosponsoring columns and rows) to be the same as the adjacency matrix of another graph, then they are the same graph. 
A: For two graphs to be isomorphic, there has to be an isomorphism between them: a bijection $f$ between their vertex sets such that there is an edge between $v$ and $w$ iff there is an edge between $f(v)$ and $f(w)$.  So to show they are isomorphic, you have to write down such a bijection.  The conditions you are saying are consequences of the existence of an isomorphism, but they might happen to hold coincidentally even if there does not exist an isomorphism.
