For a commutative group scheme $\pi \colon G \to S$ finite locally free over a base scheme $S$ we let $A := \pi_* \mathcal{O}_G$ and $A^* = \mathcal{Hom}_\mathcal{O_S}(A, \mathcal{O}_S)$. Then $A^*$ ist again a finite locally free $\mathcal{O}_S$-Algebra and the Cartier dual to $G$ is $G^* := Spec(A^*)$.
I want to understand the fact, that $G^*$ represents the inner hom $T \mapsto Hom_{Grp/T}(G_T, \mathbb{G}_{m,T})$ in the category of $S$-schemes. I already have bijections $$ Hom_S(S, G^*) \cong \{ x \in \Gamma(S, A)^\times \mid \Delta(x) = x \otimes x, \epsilon(x) = 1\} \cong Hom_{Grp/S}(G, \mathbb{G}_{m,S}) $$
where $\Delta$ and $\epsilon$ denote co-multiplication and augmentation of $A$ respectively. I got two questions:
I think that $(G^*)_T \cong (G_T)^*$ canonically, but I am not quite sure why this is true. It would however suffice to obtain $Hom_S(T, G^*) \cong Hom_{Grp/T}(G_T, \mathbb{G}_{m,T})$ for any $S$-scheme $T$.
I have absolutely no idea how to show this bijection to be functorial. I have tried to simply chase elements in the significant diagrams, but with all the isomorphisms I have used to obtain the bijection, it just seems impenetrable to me. Is there a way to reduce to a situation where the functoriality is easier to see?
For 1. I think you can proceed as follows. You want to understand $(G^\ast)_T$, but this is just $\underline{\mathrm{Spec}}(f^\ast A^\ast)$ where $f:T\to S$. But,
$$f^\ast A^\ast=\mathcal{Hom}_{\mathcal{O}_T}(f^\ast A,\mathcal{O}_T)$$
and
$$f^\ast A=f^\ast\pi_\ast\mathcal{O}_G=(\pi_T)_\ast(\mathrm{pr}_G^\ast(\mathcal{O}_G))=(\pi_T)_\ast(\mathcal{O}_{G_T})$$
(note I used the fact that $\pi$ was locally free here, in particular, I'm using a baby case of Grauert's theorem). Thus,
$$f^\ast A^\ast=\mathcal{Hom}_{\mathcal{O}_T}(f^\ast A,\mathcal{O}T)=\mathcal{Hom}_{\mathcal{O}_T}((\pi_T)_\ast(\mathcal{O}_{G_T}),\mathcal{O}_T)$$
and so clearly
$$(G^\ast)_T=\underline{\mathrm{Spec}}(f^\ast A^\ast)=\underline{\mathrm{Spec}}(\mathcal{Hom}_{\mathcal{O}_T}((\pi_T)_\ast\mathcal{O}_{G_T},\mathcal{O}_T))=(G_T)^\ast$$
For 2., you can reduce to the case when $S$ is affine, and then I think your question is answered here. Please let me know if I am mistaken!