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By def- inition, if G′ is such a graph isomorphic to G, there exists a bijection f:V →V that is an isomorphism of G and G′.The number of all possible bijections f : V → V is n!, and hence G is isomorphic to at most n! distinct graphs on the set V . (We may be overcounting! For our specific example for n = 3, we had 3! = 6 bijections but only 3 distinct graphs isomorphic to G—can you explain why?)

In other words, each class of the equivalence $\cong$ on the set of all graphs with vertex set V consists of no more than n! graphs, Therefore, the number of equivalence classes is at least

$$\frac{2^{\frac{n}{2}}}{n!}$$

My question 1 is on the first highlighted part, is the reason because every edge can be counted twice therefore, the 3 more? like if the same graph in a mirror?

my second question is I still don't fully understand the concept of equivalence on the second highlight.

Please show me some examples with detail explaination. Thanks.

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    We don't have your "specific example" -- but to get some intuition, consider the complete graph $K_n$ on $n$ vertices. Do any permutation of the vertices: you still have the same grap, $K_n$. You have $n!$ permutations of the vertices, but they all transform $K_n$ into itself. There is only one graph isomorphic to $K_n$: $K_n$ itself. – 2017-02-02
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    @ Clement C. The example that I forgot to type is {1,2,3}. – 2017-02-03
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    This is the set of vertices $V $, not the full graph. – 2017-02-03

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