Let $S$ be a scheme.
I want to show that the composition $\text{Spec}_S(-)\circ \text{Sym}(-)$ is a full and faithful contravariant functor from the category of quasi-coherent $\mathcal{O}_S$-modules (morphisms are morphisms of $\mathcal{O}_S$-modules) to the category of $S$-schemes that are isomorphic to $\text{Spec}_S(X)$ for some $S$-scheme $X$. Here, $\text{Spec}_S(-)$ is relative Spec and $\text{Sym}(-)$ is the symmetric algebra.
Now, let $E$ and $F$ be quasi-coherent $\mathcal{O}_S$-modules. Then, noting that $\text{Spec}_S(-)$ is fully-faithful and using the universal property of $\text{Sym}(-)$, we have \begin{align*} \text{Hom}_{S-\text{schemes}}(\text{Spec}_S(\text{Sym}(E)),\text{Spec}_S(\text{Sym}(F))) &= \text{Hom}_{\mathcal{O}_S-\text{algebras}}(\text{Sym}(F)),\text{Sym}(E))\\\ &= \text{Hom}_{\mathcal{O}_S-\text{modules}}(F,\text{Sym}(E)) \\\ \end{align*}
I want to show that the last expression is $\text{Hom}_{\mathcal{O}_S-\text{modules}}(F,E)$, but I'm not convinced this should be true (this isn't exactly the same, but it's not true that $\text{Hom}_{A-\text{modules}}(B,C)$ is the same as $\text{Hom}_{A-\text{modules}}(B,\text{Sym}(C))$ if $B,C$ are $A$-modules).
What should be fixed/where did I go wrong? Thank you!
Question: "What should be fixed/where did I go wrong? Thank you!"
Answer: There is no such isomorphism for the following reason: Assume $E:=A\{x_1,..,x_n\}$ and $F:=A\{y_1,..,y_m\}$ are free $A$-modules of rank $n,m$. It follows
$$Sym_A^*(E)\cong A[x_1,..,x_n], Sym_A^*(F)\cong A[y_1,..,y_m]$$
are polynomial rings. Hence
$$Hom_{A-alg}(Sym_A^*(F), Sym_A^*(E)) \cong Hom_{A-mod}(F, Sym_A^*(E)) \cong Sym_A^*(E)^m$$
and
$$Hom_{A-mod}(F,E) \cong A^{m \times n}$$