There is a much more general story here, though one my brain is not very up to explaining it this afternoon, and unfortunately, I don't know of anywhere it's summarized well for beginners.
So, how you you prove the restriction rule you mentioned above? You note that the restriction of a S_n rep to an S_{n-1} rep has an action of the Jucys-Murphy element X_nwhich commutes with S_{n-1}. The different S_{n-1} representations are the different eigenspaces of the J-M element.
So, one can think of "restrict and take the m-eigenspace" as a functor E_m; this defines a direct sum decomposition of the functor of restriction.
Of course, this functor has an adjoint: I think the best way to think about this is as F_m=(k[S_n]/(X_n-m)) \otimes_{k[S_{n-1}]} V http://latex.mathoverflow.net/png?F%5Fm%3D%28k%5BS%5Fn%5D%2F%28X%5Fn%2Dm%29%29%0A%5Cotimes%5F%7Bk%5BS%5F%7Bn%2D1%7D%5D%7D%20V.
These functors E_m,F_m satsify the relations of the Serre relations for \mathfrak{sl}(\infty) http://latex.mathoverflow.net/png?%5Cmathfrak%7Bsl%7D%28%5Cinfty%29. Over characteristic 0, these are all different, and you can think of this as an \mathfrak{sl}(\infty) http://latex.mathoverflow.net/png?%5Cmathfrak%7Bsl%7D%28%5Cinfty%29. If instead, you take representations over characteristic p, ythen E_m=E_{m+p} so you can think of them as being in a circle, an affine Dynkin diagram, so one gets an action of \widehat{\mathfrak{sl}}(p) http://latex.mathoverflow.net/png?%5Cwidehat%7B%5Cmathfrak%7Bsl%7D%7D%28p%29.
Similar categorifications of other representations can deconstructed in general by looking at representations of complex reflection groups given by the wreath product of the symmetric group with a cyclic group. So, Sammy, you shouldn't rescale, you should celebrate that you found a representation with a different highest weight (also, if you really care, you should go talk to Jon Brundan or Sasha Kleshchev; they are some of the world's experts on this stuff).
EDIT: Khovanov has actually just posted a paper which I think might be relevant to your question.