Let $S$ be a scheme and let $G$ be a group scheme over $S$.
Let $X$ be an $S$-subscheme of $G$; we can define the controvariant functor: $$ \mathbf{N}_G(X):\mathbf{Sch}_{S}\to\mathbf{Group}\\ \forall T,U\in Ob(\mathbf{Sch}_{/S}),\,\varphi\in\hom_{\mathbf{Sch}_{/S}}(T,U),\\ \mathbf{N}_G(X)(T)=\{g\in G(T)\mid \forall t\in X(T),\,g^{-1}tg=t\}=N_{G(T)}(X(T)),\\ \mathbf{N}_G(X)(\varphi):g\in N_{G(U)}(X(U))\to g\circ\varphi\in N_{G(T)}(X(T)), $$ and if it is representable by an $S$-subscheme $N_G(X)$ of $G$ we state that $N_G(X)$ is the normalizer of $X$ in $G$.
In a similar way, one can define the stabilizer functor $\mathbf{Stab}_G^{\alpha}(X)$ with respect to an action $\alpha$ of $G$ on $X$; and one can prove that this functor is ever representable by an $S$-subgroup scheme $Stab_G^{\alpha}(X)$.
These constructions are clear, for me.
I want define the conjugation action $\gamma$ of $G$ on itself; my idea is declare that for all $S$-scheme $T$, $\gamma(T):G(T)\times G(T)\to G(T)$ is the conjugation action of $G(T)$ on $G(T)$. In consequence, one has that $N_G(X)=Stab_G^{\gamma}(X)$.
My question is: exists an $S$-group scheme $G$ such that this idea doesn't work?
I hope that my question is (sufficiently) clear.
Edit: I (hopefully) have clarify the question, as asked me from David Loeffler.
Edit 2: No, $\gamma$ exists ever; and moreover: if $x$ is a valued point of $G$ then $\mathbf{Stab}_G^{\alpha}(x)$ is ever representable.
I change my question: exists $G$ such that $\mathbf{N}_G(H)$ is not representable, for some $S$-subgroup scheme $H$?