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Let $\Gamma =K_{3,3}$ be the the complete bipartite graph on $3$ vertices and let $G=\text{Aut}(\Gamma)$ be its symmetry group. Suppose we wish to determine the size of $G$.

Here is what I think,

We let $G$ act on the vertices of $\Gamma$ and consider the labelling $\{1,2,3,4,5,6\}$ going from top left down and then top right down.

Clearly the action is transitive since we can move $1$ to $2$ and $3$ by simply permuting the left hand vertices and we can also compose this with a reflection down the middle to get $1$ to $4,5,6$.

Hence by the orbit stabiliser theorem we get $|O_1|=|G|/|\text{Stab}_G(1)|$ I feel like the only elements of $G$ in the stabiliser of $1$ will be the identity, the map that switches $2$ and $3$ and any permutation of the right hand side. So in total this gives $2 \times 3!=12$ elements in the stabiliser.

So overall $|G|=6 \times 12=72$?

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    Yes that's right. I would probably calculate it as $3!$ permutations of each of the blocks, and you can also swap the blocks, making a total of $3! \times 3! \times 2 = 72$. This kind of group is known as a *wreath product*. – 2017-01-27

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