You can certainly have non-equivalent monadic functors. Here's one example: Let $\mathcal{V}_k$ be the category of $k$-vector spaces. For a vector space $V$, let $H_V: \mathcal{V}_k\to \mathcal{V}_k$ be the functor
$$
H_V(W) = hom_k(V,W).
$$
Such a functor is always monadic, as long as $V$ is non-zero and finite dimensional. The associated monad is
$$
T_V(W) = hom_k(V, V\otimes_k W) = End_k(V)\otimes_k W,
$$
so this is presenting a Morita equivalence: $k$-vector spaces are equivalent to modules over the matrix ring $End_k(V)$.

You wanted functors to set; let $U_V:\mathcal{V}_k\to Set$ be given by the same formula as $H_V$. 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 $H_V$ is an exact functor, so the hypotheses of the Barr-Beck theorem certainly hold, though $T_V$ is not tensoring with an endomorphism ring if $V$ is infinite.).