Rearrangement with sodium iodide (ring opening)
I think the final product will also contain some Bromine atoms attached to it. and the no. of double bonds will be one less than that of the compound given in the picture. I may be wrong but with all of my known possibility of Organic reactions, I can actually reach to the final product somewhat different from that given in the picture  and I don't think the final product given in the picture can ever be got from the given starting material. I will be happy if someone else can find flaws in the logic  given below.
 
$\ce{NaI}$ in acetone is the reagent used for Finkelstein reaction which is nothing but a standard substitution reaction with $\ce{I-}$ ions. But in presence of vicinal dibromides, or any special structure where there is a chance of elimination of hallide ions even further, the iodide ions can actually make the molecule undergo preferably an $E_2$ elimination (in most of the cases).


In this starting compound also, there are substitutable $\ce{Br}$ atoms present. So, the iodide ions first performs a substitution reaction, and then undergo an elimination via bond migration. This same procedure occurs for two times. and finally we are left with a product which has $\ce{Br}$ atoms attached to $sp^2$ carbon atoms, and therefore they are highly reluctant to undergo any substitution or elimination reaction. So, that's why, logically the end product should be, $\text{ (1E,3Z,5E,7E,9Z,11E)-2,5,8,11-tetrabromo-6a,12a-dihydrooctalene}$. Here is the structure of the possible end product. 



$\textbf{Mechanism:-}$



As far as I see, there can not be any more further reaction to the final compound. According to me, the given compound in the picture should not be the final product if we take the starting material to be the correct one. Instead, the end product in the mechanism should be the one produced with higher yield at the end.