Prove that set $A$ a finite abelian group and $B$ multiplicative group of all homomorphisms from $A$ to units of $C$ are isomorphic 
Let $A$ be a finite abelian group and let $B := \{f: A \to \mathbb  C^{×} | f$ is a group homomorphism $\}$. It can be easily checked that $B$ is an abelian group via $fg(x) = f(x)g(x)$.
  Prove that $A$ is isomorphic to $B$. Also, prove that if $f$ is not an identity element of $B$ then $\sum _{a \in A} f(a) = 0$.

This problem is from a module theory class which previously dealt with group theory. I have no idea how this would relate to module theory.
I attempted to look at a special case, $A = \mathbb{Z / 2Z}$. For notational simplicity denote the two elements by $1,0$. To find a corresponding isomorphism, we must send $1$ to some $f$. say $f(0) = a, f(1) = b$. Then $a = f(0) = f(2 * 1) = 2f(1) = 2b$, and similiarly $a = 4b$, so that $a = b = 0$, which is inpossible since the image of $f$ must lie outside of $0$.
Am I doing something wrong here? Any input for solving the problem - not just critic of my attempt - would be greatly appreciated. The second part, which I tried to prove by assuming the first, didn't work out so well. How is module theory related to this?

The problem with your special case is that we want the multiplicative group of the units of $\mathbb{C}$, so we have that $f(2 \cdot 1) = f(1)^2$, not $f(2 \cdot 1) = 2f(1)$. Since $f$ is a homomorphism, it maps the identity element of $A$ to the identity element of $\mathbb{C}^\times$, and so $f(0) = 1$. Since $f(1)^2 = f(0) = 1$, we get that $f(1) = -1$ or $f(1) = 1$.
In the general case, the classification of finite abelian groups tells us that
$$
  A \simeq \mathbb{Z} / n_1 \mathbb{Z} \times \mathbb{Z} / n_2 \mathbb{Z} \times \cdots \times \mathbb{Z} / n_k \mathbb{Z} 
$$
for some natural numbers $n_1, n_2, \dots, n_k$.
The homomorphism $f : A \to \mathbb{C}^\times$ is then uniquely determined by the images of the elements $\eta_1 = (1, 0, 0, \dots, 0), \eta_2 = (0, 1, 0, \dots, 0), \dots, \eta_k = (0, 0, \dots, 0, 1)$.
Since $n_i \eta_i = 0$ for each $i$, we have that $f(\eta_i)^{n_i} = f(0) = 1$, and so $f(\eta_i)$ is some $n_i^\text{th}$ root of unity.
Let $\zeta_i$ be a primitive $n_i^\text{th}$ root of unity.
To construct the isomorphism between $A$ and the group of homomorphisms from $A$ to $\mathbb{C}^\times$, we can map each element $(a_1, a_2, \dots, a_k)$ of $A$ to the function which maps $\eta_i$ to $\zeta_i^{a_i}$ for each $i$. We then just need to check that this does indeed define an isomorphism.
As for the final problem, let $f$ be some non-identity element of $B$. Then $f(\eta_i) \neq 1$ for some $i$. Suppose without loss of generality that $f(\eta_1) \neq 1$. We then have that
$$
  \sum_{a \in A} f(a) = \sum_{a_1=0}^{n_1-1} \sum_{a_2=0}^{n_2-1} \cdots \sum_{a_k=0}^{n_k-1} f((a_1, a_2, \dots, a_k)) = \sum_{a_1=0}^{n_1-1} \sum_{a_2=0}^{n_2-1} \cdots \sum_{a_k=0}^{n_k-1} f((a_1, 0, \dots, 0)) f((0, a_2, \dots, a_k)) = \sum_{a_1=0}^{n_1-1} f((1, 0, \dots, 0))^{a_1} \sum_{a_2=0}^{n_2-1} \cdots \sum_{a_k=0}^{n_k-1} f((0, a_2, \dots, a_k)).
$$
But $f((1, 0, \dots, 0))$ is a $n_1^\text{th}$ root of unity which is not equal to $1$, and for any $n_1^\text{th}$ root of unity $\omega \neq 1$, we have that
$$
  \sum_{a_1=0}^{n_1-1} \omega^{a_1} = 0,
$$
and so we have that
$$
  \sum_{a \in A} f(a) = 0.
$$
