You can certainly have non-equivalent monadic functors. Here's one example: Let mathcalVk be the category of k-vector spaces. For a vector space V, let HV:mathcalVktomathcalVk be the functor
HV(W)=homk(V,W).
Such a functor is always monadic, as long as V is non-zero and finite dimensional. The associated monad is
TV(W)=homk(V,VotimeskW)=Endk(V)otimeskW,
so this is presenting a Morita equivalence: k-vector spaces are equivalent to modules over the matrix ring Endk(V).
You wanted functors to set; let UV:mathcalVktoSet be given by the same formula as HV. Then again, this will be monadic, as long as V is non-zero and finite dimensional (and I'm not sure you even really need the finite dimensionality condition for either of these examples; added: you certainly don't in the first example, since HV is an exact functor, so the hypotheses of the Barr-Beck theorem certainly hold, though TV is not tensoring with an endomorphism ring if V is infinite.).
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