Connection Between Automorphism Groups of a Graph and its Line Graph First, the specific case I'm trying to handle is this:
I have the graph $\Gamma = K_{4,4}$.
I understand that its automorphism group is the wreath product of $S_4 \wr S_2$ and thus it is a group of order 24*24*2=1152.
My goal is to find the order of the AUTOMORPHISM GROUP of the Line Graph: $L(\Gamma)$.
That is - $|Aut(L(G))|$
I used GAP and I already know that the answer is 4608, which just happens to be 4*1152.
I guess this isn't a coincidence. Is there some sort of an argument which can give me this result theoretically? 
Also, I would use this thread to ask about information of this problem in general (Connection Between Automorphism Groups of a Graph and its Line Graph).
I suppose that there is no general case theorem.
I was told by one of the professors in my department that "for a lot of cases, there is a general rule of thumb that works" although no more details were supplied.
If anyone has an idea what he was referring to, I'd be happy to know.
Thanks in advance,
Lost_DM

I'm not sure if this is what you are asking, but there is always a map from $\mbox{Aut}(G)$ to $\mbox{Aut}(L(G))$ that is an injection in your case, explaining your divisibility relation. If $\phi$ is an automorphism of $G$, then $\phi$ also acts on $L(G)$: $\phi$ already knows what to do to vertices of $L(G)$ (which are edges of $G$), and $\phi$ also preserves the edge incidence relation, so extends to a map on edges of $L(G)$. This is not always an injection: consider the case of the graph with a single edge connecting distinct vertices. But in your case the map is an injection, which explains your divisibility relation.
