Suppose $X$ is a graph with a set $V$ of vertices and $|V|=n$. $I_X$ is the isomorphy class of $X$ and $\text{Aut}(X)$ is the automorphism group of $X$.
How can I prove the formula
$$ |I_X|=\frac{n!}{|\text{Aut}(X)|}?$$
My motivation is to show with the formula above that a graph is asymmetrical if and only if it has $n!$ isomorphic graphs. Then I would have that $|I_X|=n!$ and it would follow $|\text{Aut}(X)|=1$ and I would be ready. Maybe there is an easyier way without using this formula.
On
You have the group action of $S_n$ acting on the set of all graphs with $n$ vertices by permuting the vertex labels. I believe the orbit-stabilizer theorem is then what you want: http://en.wikipedia.org/wiki/Group_action#Orbit-stabilizer_theorem_and_Burnside.27s_lemma
$S_n$ acts on the set of all graphs on vertex set $\{1,\ldots,n\}$. In this action, the orbit containing $X$ has size $|I_X|$, and the set of elements in $S_n$ that fixes the edge set of $X$ setwise has size $|Aut(X)|$. By the orbit-stabilizer lemma, $|S_n|=|I_X|~|Aut(X)|$.