Yeah, 2,3-dimethyl-3-pentanol was actually my first idea for the answer, but a friend of mine working on it convinced me it couldn't be since there weren't three equivalent methyl groups in that structure (i.e. 9 equivalent hydrogens).
As far as leniency to the rules of NMR, I'm not too well versed, so when it comes to something like the possibility of having the 3H of carbon 1 and the 3H of the first methyl group exist virtually as the same peak (at virtually the same chemical shift) as the 3H on carbon 5, I wasn't sure how likely that was. It does make sense that, assuming there is room for this type of thing, they would all show up at about the same spot, since the are virtually identical except that there is no methyl group off of carbon 4 to provide total symmetry.
So I assume the majority of you are thinking it's 2,3-dimethyl-3-pentanol? Because now I'm very tempted to revert back to that answer too.