I'll motivate my question with a neat observation: let $G$ be a group, and let $\mathcal{P}(G)$ denote the power set of $G$, regarded as a category in the usual way (i.e., as a poset converted into a categry). If $S\subseteq T$ are subsets of $G$, then the centralizers of $S$ and $T$ satisfy $C_G(T)\subseteq C_G(S)$. As it turns out, the centralizer determines a functor: $$ C_G:\mathcal{P}(G)^{\mathrm{op}}\to\mathcal{P}(G),\quad C_G:S\mapsto C_G(S). $$ One can show that $C_G$ is a right adjoint to its own opposite $C_G^{\mathrm{op}}:\mathcal{P}(G)\to\mathcal{P}(G)^{\mathrm{op}}$. We therefore have a "double centralizer monad" $T = C_G^2 = C_G\circ C_G^{\mathrm{op}}$ induced from this adjunction (the algebras are the "self-bicommutant" subgroups of $G$).
With this in mind, my question is:
Q. Can the normalizer be viewed in a similar way?
Modifications to the category in question would likely be needed, since normalizers don't preserve (or reverse) inclusions...
As Arturo described, this property of the centralizer comes from the Galois connection. So I think it would be unlikely to find a similar setting for the normalizer.
Despite this, I'll give a different categorical description of the normalizer so that this question is resolved.
Let $\mathcal{P}(G)$ be as above, and let $G$ act on it via conjugation. Then, we may consider the action groupoid $\mathcal{P}(G)//G$: this has subsets of $G$ as the objects, and morphisms from $S$ to $T$ given by elements $g\in G$ such that $g\cdot S = gSg^{-1} = T$.
Then, the automorphism group $\mathrm{Aut}_{\mathcal{P}(G)//G}(S)$ of an object $S$ in this category is the stabilizer of $S$ under this action, which is precisely $N_G(S)$.