If $G$ acts on $A$ faithfully/freely/transitively, what can we say about its action on $\hat A$? Let $A$ be a finite abelian group and let $G\subset \operatorname{Aut}A$, with $G$ acting from the left. $G$ acts in a natural way (on the left) on $A$'s character group $\hat A$ by $g\chi = \chi\circ g^{-1}$. My question is, if we know something about $G$'s action on $A$, can we say anything about its action on $\hat A$? More specifically:

(1) If $G$'s action on $A\setminus\{0\}$ is free, is its action on $\hat A\setminus\{0\}$ free? (2) $G$'s action on $A$ is faithful by construction; is its action on $\hat A$ faithful? (3) If its action on $A\setminus\{0\}$ is transitive, is its action on $\hat A\setminus\{0\}$ transitive?

My intuition says yes to all 3 questions, but I really do not see the argument so I have no idea whether to trust it. I think it is based on the fact that (a) $A$ and $\hat A$ are isomorphic (though not canonically), and (b) $A$ is canonically isomorphic to $\hat{\hat{A}}$, and the canonical isomorphism commutes with the action of $G$. But logically these don't imply anything. I can't extract any info from (a) unless one of the isomorphisms $A\rightarrow\hat A$ commutes with the action of $G$ on each, but usually none of them do. And (b) has nothing direct to say about $\hat A$ or $G$'s action on it.
It would be enough to conclude (1), (2) and (3) if I could guarantee that there is an isomorphism $A\rightarrow\hat A$ that commutes with the action of $G$ up to twisting $G$ by some automorphism. This seems potentially plausible to me but I don't see the argument; it's true in the one case I've worked out by hand ($A=C_7$, $G=C_6 = \operatorname{Aut} A$), but there are lots of special reasons why it should be true in that case.
 A: This is a partial answer, answering questions (1) and (2) but not (3). The methods are ad-hoc. I remain interested in a more unified approach, and of course I'd also like to know the answer to (3).
The answer to question (2) is yes. By basic representation theory, the characters of $A$ separate the elements of $A$. Thus an automorphism $g$ of $A$ can't fix every character without also fixing every element. Thus the action of $G$ on $\hat A$ is faithful.
The answer to question (1) is yes. I argue combinatorially:
We assume $G$'s action on $A\setminus\{0\}$ is free, which implies the same is true of the restriction of the action to any subgroup of $G$, and in particular of the cyclic subgroup generated by any element $g\in G$. 
The statement that $G$'s action on $\hat A\setminus \{0\}$ is free is equivalent to the statement that for any nontrivial character $\chi\in \hat A$, $g\chi = \chi$ implies $g=1$. Now any nontrivial character $\chi$ is a homomorphism to a cyclic group of order $r\geq 2$. The fibers of this homomorphism all have cardinality $m = |A|/r$. One of them contains $0$ and thus it contains $m - 1$ nonidentity elements, and there is also at least one other fiber, with $m$ nonidentity elements.
Suppose $g$ satisfies $g\chi = \chi\circ g^{-1} = \chi$. Then $\chi\circ g^{-1}$ has the same fibers as $\chi$, and it follows that $g^{-1}$, and thus $g$, acts separately on each fiber. Thus each fiber is a union of orbits for the action of $\langle g\rangle$ on $A$.
As observed above, the action of $\langle g\rangle$ on $A\setminus\{0\}$ is free, which means that all orbits for $\langle g\rangle$'s action on $A\setminus\{0\}$ have the same length, namely the order of $g$ (let's call it $d$). Thus $m$ (the cardinality of the fibers not containing the identity, of which as noted above there is at least one) is a multiple of $d$. But $m-1$ is also a multiple of $d$, since this is the cardinality of the part of $A\setminus\{0\}$ sitting in the fiber containing $0$.
Thus $d\mid (m,m-1)$, i.e. $d=1$, i.e. $g$ is order $1$, i.e. $g=1$. This proves the action of $G$ on $\hat A$ is free.
