EDIT: The argument does not work.
Answer to 1: No.
Let $\phi:$ Sym($M$) $\to$ Sym($N$) be an $A$-algebra isomorphism. We will see that it induces an $A$-module isomorphism $\tilde \phi: M \to N$. Pick a set of $A$-module generators $m_1, \ldots, m_k$ of $M$. These also generate Sym($M$) as an $A$-algebra. Let $\phi(m_i) = \sum_j n_{ij}$ with $n_{ij} \in N^j$. In particular each $n_{i1} \in N$. I claim that $\tilde \phi: m_i \to n_{i1}$ gives a well defined $A$-module map from $M$ to $N$. To see it, all you have to show (I think) is that if $\sum a_im_i = 0$ for $a_1, \ldots, a_k \in A$, then $\sum a_in_{i1} = 0$. But it is true, because $\phi$ is an $A$-algebra homomorphism and hence $0 = \phi(\sum a_i m_i) = \sum_j (\sum_i a_i n_{ij})$ and thus the inner sum is $0$ for each $j$, because it is the $j$-th graded component. The claim is proved.
In the same way you can show $\phi^{-1}$ also induces a map $N \to M$ and it should be inverse to $\tilde \phi$.
This argument seems to show that there is a map from $Hom_{A-alg}(Sym(M),Sym(N))$ to $Hom_{A-mod}(M,N)$. But I could not have seen it before writing it out.