For an equal mass, the feathers will have a larger volume and therefore more buoyancy, so the [net/apparent] weight will be less
That only works if you're comparing figures measured suspended in different fluids, or in fluids which have differing diffusion rates into the materials you're weighing. for example, if I'm comparing the weight of an object suspended in air with the weight of the same object suspended in a fluid of density 8.33 pounds per gallon (US) (which happens to be the density of normal salinity sea-water at 70 deg F, though why such an insane collection of fucked-up units is the norm in my business is just a deranged accident of history), then I'd get a buoyancy force which is density dependent. But if I'm comparing the weights of two objects of the same density (say, feathers and 1mm nylon filament) suspended in the same fluid, then the correction made for buoyant forces is the same in both cases. (If the bulk density of the keratin that comprises feathers is the same as that of nylon, which I don't know for sure, but it probably isn't far off.
If you have a medium of changing density - e.g. air that is warming over time - then you get lots of complicated issues of the permeability of the stationary medium with respect to the fluid medium and a whole crapload of other non-equilibrium effects.
Buoyancy forces per se have almost nothing to do with the volume occupied by a bag of a medium (e.g. the bag of feathers you seem to be thinking of) because that bag is a mixed medium - it contains feathers and trapped air. Unless you're taking steps to prevent the air from being displaced by your new fluid medium (e;g; a waterproof bag) then when things equilibrate, you'll be back to considering buoyancy forces controlled by the relative bulk density of the materials involved.
Perhaps considering a less extremely misleading example might help to clarify matters. I'm at sea (8.33 pounds per gallon [ppg]) above an oil well filled with drilling fluid of 10.2ppg density. If I assemble a string of tools made out of steel (density 7.something tonnes per cubic metre - I'd have to look it up, because steels vary) then they'd weigh something in air, something different if I lowered them into the seawater (as if I were starting the well, but had nothing in place to hold my denser drilling fluid in place - what we call "spudding the well"), and something different again if I had a container attached to the rock of the seabed, which I'd filled with my 10.2ppg drilling fluid.
Now replace those drilling tools with titanium ones at a density of 3.something tonnes per cubic metre. Again I'd get 3 different weights in air, sea-water and my current density of drilling fluid. And those buoyancy forces are going to severely affect the amount of force I can exert on the bottom of the hole using the buoyed weight of those tools. What we call the available weight on bit. Which is a major constraint on our speed of drilling. And which is something I've just had to be double-checking for our next assembly of drilling tools.
You're confusing the complicating effects of surface trapped air with the actual buoyant forces involved. Not good physics.
By the way, if you ever worked with precision scales, you'd know that you have to - absolutely have to - take into account the buoyancy of air when you get to the 5th significant digit of your measurement - and you really ought to be doing it at the 4th significant digit. You're already well into the territory where you need to be drying your reagents before weighing, even if they're not particularly hygroscopic (absorbing moisture from ambient air).