Let $\mathfrak{2}$ be the category with two objects $0,1\in \mathfrak{2}$ and a single non-identity arrow $0\to 1$. Also, let $i_0:C\to C\times \mathfrak{2}$ be defined where $i_0(c)=(c,0)$ for all objects $c$ in the category C and let $i_1:C\to C\times \mathfrak{2}$ be defined where $i_1(c)=(c,1)$ for all objects $c$ in the category C. Fixing a parallel pair of functors $F,G:C\to D$, natural transformations $\alpha: F\Rightarrow G$ correspond bijectively to functors $H:C\times \mathfrak{2}\to D$ such that $Hi_0=F$ and $Hi_1=G$.
How do I construct a bijection to prove this?
It's simple:
Let ${\bf a} :0\to 1$ denote the nonidentity arrow.
For an object $c$ in $C$, we have $\alpha_c:Fc\to Gc$. Let $H(c, {\bf a}):=\alpha_c$, and obviously $H(c,0)=Fc$ and $H(c,1)=Gc$.
For an arrow $\varphi:c\to c'$ of $C$, define $H(\varphi,{\bf a})$ as the diagonal composition of the commutative square of $\alpha$.
For the converse, if $H$ is given, you can actually obtain $F$ and $G$ as well, by the given formulas, and simply define $\alpha_c:=H(c, {\bf a})$.