Construct group isomorphism $\alpha : \mathbb{Z}^n \rightarrow \mathbb{Z}^n$ s.t. $g=f \circ \alpha$? Suppose we have surjective group morphisms
$$f: \mathbb{Z}^n\rightarrow A \qquad g:\mathbb{Z}^n\rightarrow A.$$
How do I construct a group isomorphism $\alpha:\mathbb{Z}^n \rightarrow \mathbb{Z}^n$ such that $g=f \circ \alpha$ ? 
 A: Apparently I don't have the privilege of adding a comment yet, but I would like to point out that Dylan's answer is correct since a homomorphism on a free abelian group is determined by where it sends the basis.  HOWEVER the original problem has no general solution since in general $\alpha$ will not be an isomorphism.
For example, if $n=1$, $A=\mathbb{Z}/5\mathbb{Z}$, $f(1)=[1]$ and $g(1)=[2]$ (surjective since $2$ and $5$ are coprime), then there is no isomorphism $\alpha:\mathbb{Z}\rightarrow\mathbb{Z}$ such that $g=f\circ\alpha$
A: This was way too long for a comment so I'm posting this here. Don't consider this as a complete answer ; I am saying no such thing. I am only saying that the problem is not as general as it seems.
$f : \mathbb Z^n \to A$ is surjective hence $\mathbb Z^n / \mathrm{Ker} \, f \cong A$. For the same reasons, $\mathbb Z^n/ \mathrm{Ker} \, g \cong A$, hence 
$$
\mathbb Z^n / \mathrm{Ker} \, f \cong \mathbb Z^n / \mathrm{Ker} \, g.
$$
Now all this being said, we have the following commutative diagram :
$$
\begin{matrix}
\mathbb Z^n & \overset{\alpha}{\longleftarrow} & \mathbb Z^n \\\
\downarrow f & & \downarrow g \\\
\mathbb Z^n/ \mathrm{Ker} f & \overset{\varphi_1^{-1} \circ i \circ \varphi_2}{\longleftarrow}& \mathbb Z^n/ \mathrm{Ker} g \\\
\downarrow \varphi_1 & & \downarrow \varphi_2 \\\
A & \overset{i}{\longleftarrow} & A \\\
\end{matrix}
$$
where $\varphi_1$, $\varphi_2$ are isomorphisms and $i$ is the inclusion map. The question is whether $\alpha$ fits in there as an isomorphism (i.e. does it exist). Well, we can reduce ourselves to the study of the map $\psi = i \circ \varphi_2 \circ g$ and look at the diagram 
$$
\begin{matrix}
\mathbb Z^n & \overset{\psi}{\longrightarrow} & A \\\
  &  & \uparrow \varphi_1 \\\
 & & \mathbb Z^n / \mathrm{Ker} f\\\
\end{matrix} 
$$ 
and ask ourselves if this diagram commutes. Since $\varphi_1$ is an isomorphism it is injective and $\psi$ is surjective, therefore we can complete this diagram with a unique morphism $\Phi$ such that $\psi = \varphi_1 \circ \Phi$. Now I don't know how to TeX this but that gives us $\Phi$ going from the topright $\mathbb Z^n$ to the middle left $\mathbb Z^n / \mathrm{Ker} f$, and I've shown that this map is unique. All we need to do now is complete the diagram
$$
\begin{matrix}
\mathbb Z^n & \overset{\Phi}{\longrightarrow} & \mathbb Z^n / \mathrm{Ker} f \\\
& & \uparrow f \\\
& & \mathbb Z^n \\\
\end{matrix}
$$
so that in other words, I've reduced the problem for arbitrary $A$ to the problem of dealing with some quotient of $\mathbb Z^n$ ; given $f : \mathbb Z^n \to \mathbb Z^n$ and $\Phi : \mathbb Z^n \to \mathbb Z^n / \mathrm{Ker} f$, one completes this diagram if and only if one completes the diagram above for arbitrary $f$,$g$. Since there is no obvious reason why this diagram should be completed I expect counter examples.
