Natural transformation between $\mathcal{F}$ and $\mathcal{G}$ 
Consider the following functors from the category of finite Abelian groups $F AbGrps$ to the category of sets defined as follows:

*

*$\mathcal{F}: F AbGrps \to Sets$, a group $G$ is mapped to the set elements of order $1$ or $2$.


*$\mathcal{G}: FAbGrps  \to Sets$, a group $G$ is mapped to the set of elements of order $1$ or $3$
Define the functors on the morphisms and show that indeed we obtain functors. What is the size of the set of natural transformations between $\mathcal{F}$ and $\mathcal{G}$?

Note that $\mathcal{F}(G)$ and $\mathcal{G}(G)$ are abelian groups. For $f:A \to B$, we define $\mathcal{F}(f)(a)=f(a)$ and similarly for $\mathcal{G}$. We find that they are functors. My question is that whether they are unique functors respect to $\mathcal{F}$ and $\mathcal{G}$ or not? If so, how can I construct the natural transformation?
I guess I should take natural transformation $\alpha: \mathcal{F} \to \mathcal{G}$ which consists of $\alpha_X: \mathcal{F}(X) \to \mathcal{G}(X)$, $\alpha_X(x) \mapsto e$.
 A: You seem to have two questions. The first seems to be the following:

Question: Are the definitions given in the question the only possible ways to define functors $\mathcal{F}$ and $\mathcal{G}$ such that $\newcommand\cF{\mathcal{F}}\newcommand\cG{\mathcal{G}}\cF A = A[2]$ and $\cG A = A[3]$? (Here the notation $A[n]$ means the $n$-torsion subgroup of the abelian group $A$.)

The answer is no. Here's an example of how to construct another functor with different maps on morphisms, but the same map of objects. 
This can be done for any category $\newcommand\cC{\mathcal{C}}\cC$. First for any  object $X$ choose an automorphism $\gamma_X$. Then we have a functor $c_{\gamma} : \cC\to \cC$ (conjugation by $\gamma$) defined by $X\mapsto X$ for every object $X$, but if $f:X\to Y$ is a morphism, then $f\mapsto \gamma_{Y}^{-1} f \gamma_X$. 
Now if we choose such $\gamma_A$ for every finite abelian group, you can define
$\cF'=\cF\circ c_\gamma$ and $\cG'=\cG\circ c_\gamma$, and by choosing the automorphisms to do nontrivial things to the $2$ and $3$ torsion for some abelian group, you get that $\cF'\ne \cF$ and $\cG'\ne \cG$. 
(We could also choose automorphisms $\gamma'_A$ for every 2-torsion abelian group $A$, and then define $\cF' = c_{\gamma'} \circ \cF$ as well, to get even more possible functors.)
The definition you've given however is the "right" one. (The intended one).
It's an interesting question as to how badly uniqueness fails. What are all the functors which have this action on objects? I'm not sure of the answer.
Second Question:
Your second question is about what the natural transformations from $\cF$ to $\cG$ are.
You appear to have already constructed one, assuming you're using $e$ to denote the identity element. 
This is in fact the only natural transformation. After all, a natural transformation is 
a map from $A[2]$ to $A[3]$ for all abelian groups $A$ natural in $A$. However, 
if $x\in A[2]$, then $2x=0$, and if $y\in A[3]$, then $3y=0$, so $2y=-y$. Thus $(-2)y=y$.
Thus if $\alpha_A : A[2]\to A[3]$, then for any $x\in A[2]$, 
$$\alpha_A(x)= (-2)\alpha_A(x) = \alpha_A(-2x) = \alpha_A(0)=0.$$
Thus for any $A$, the only map $A[2]$ to $A[3]$ is the $0$ map. This is the natural transformation you've already found.
An aside and generalization: (This is likely irrelevant to you for now, but perhaps interesting to others)

Question: In general, what are the natural transformations from $A[n]$ to $A[m]$?

Answer. Note that we have a natural isomorphism 
$$\newcommand\Hom{\operatorname{Hom}}
A[n] \simeq \Hom(\Bbb{Z}/n,A),
$$
so we're interested in natural transformations from 
$\Hom(\Bbb{Z}/n,-)$ to $\Hom(\Bbb{Z}/m,-)$. By the Yoneda lemma, the set of such natural transformations is naturally isomorphic to 
$\Hom(\Bbb{Z}/m,\Bbb{Z}/n)$, and this is isomorphic to 
$\Bbb{Z}/(n,m)$. 
Let $d=\gcd(n,m)$, $n'=n/d$, $x\in \Bbb{Z}/(n,m)$. The corresponding natural transformation to $x$ is 
the map 
$A[n]\to A[m]$ defined by 
$$a\mapsto xn'a.$$
You can check that this is well defined and makes sense.
