Catalytic iodine in synthesis of phosphonium salts
Iodine is a stronger nucleophile here than bromine


I agree with your HOMO-LUMO reasoning, but this doesn't really matter too much.  As long as iodide is a strong enough nucleophile to displace bromide at least some of the time, then the reaction we want to happen, will happen.  Of course the more nucleophilic iodide is (here's where your HOMO-LUMO argument comes into play), the faster the desired reaction will happen.

I pretty much agree with the rest of what you say Jori, bit I might say it just a bit differently.

I'm not sure how triphenyl phosphine (TPP) is hybridized, I'd guess that it is unhybridized just like phosphine ($\ce{PH3}$).  If it is unhybridized than our lone pair on phosphorous is in an s-orbital.  Since the s-orbital has little "directionality", this lone pair on phosphorous will be a weak nucleophile.  On the other hand, if TPP is hybridized (like triphenyl amine), then although the lone pair is now in an orbital with high p-character (and therefore very directional) it will be extensively delocalized into the 3 aromatic rings (just like triphenyl amine which is a very weak base for just that reason), which again means that the lone pair isn't very available and consequently TPP will be a weak nucleophile.  So now matter how TPP is hybridized, we expect it to be a weak nucleophile.

Iodide is, as you note, a much better leaving group then bromide.  In fact, iodide is such a good leaving group that it can be readily displaced by even a poor nucleophile like TPP. Iodide is also a good nucleophile and can displace bromide.

The reaction is run with just a small (catalytic) amount of iodide.  As soon as iodide replaces a bromide (nucleophilic displacement), the iodide, being such a good leaving group, will now be quickly displaced by the weakly nucleophilic TPP.  During this last step, iodide is regenerated and the cycle starts again.