Recall the definition of a natural transformation: Let $F,G : C \to D$. Then an assignment of $\alpha_c : F c \to G c$, for each $c\in C$, is natural if, for every $f : c \to c'$ in $C$, we have that the following commutes:
$$
\begin{array}
(Fc & \xrightarrow{\alpha_c} & G c \\\
\downarrow{Ff} & & \downarrow{Gf} \\\
Fc' & \xrightarrow{\alpha_c'} & Gc'
\end{array}
$$
TLDR: In nice categories, the naturality condition requires that any choice made to define $\alpha_c$ and $\alpha_{c'}$ is preserved by $Ff$ and $Gf$, for every $f: c \to c'$. This does not happen when we make arbitary (incompatible) choices for $c$ and $c'$. See end remarks for a full answer to your question.
Now, suppose, for example, that $C=D=Vect_\mathbb{K}$, and $F=G=Id_{Vect_\mathbb{K}}$. There is of course the identity natural isomorphism, with each $\alpha_c$ the identity linear transformation. Now, what would happen if we try to define another natural isomorphism by making an arbitrary choice of basis in order to define any $\alpha_c$? Say the basis we choose for $Fc=c$ is $\{e_1,...,e_n\}$ and we defined $\alpha_c$ by sending this basis to another basis $\{v_1,...,v_n\}$. But, for $Fc'=c'$ we might have made the choice of basis of $\{e'_1,...,e'_m\}$ and defined $\alpha_c'$ by sending this basis to another basis $\{w_1,...,w_m\}$. Then each component is indeed an isomorphism since it is defined by a bijection of basis.
Now, what would the naturality condition would require? We would need the following diagram to commute:
$$
\begin{array}
(c & \xrightarrow{\alpha_c} & c \\\
\downarrow{f} & & \downarrow{f} \\\
c' & \xrightarrow{\alpha_c'} & c'
\end{array}
$$
Chasing the basis of $c$ along the top-right route lends $\{f(v_1),...,f(v_n)\}$. Chasing the basis of $c$ along the left-bottom route lends $\{\alpha_{c'}(f(e_1)),...,\alpha_{c'}(f(e_n))\}$. So naturality would require that $f(v_i)=\alpha_{c'}(f(e_i))$, for ever linear transformation $f:c\to c'$.
But there is no reason to expect this to happen for every $c, c'$ and every linear transformation $f$, since choice of basis is not necessarily preserved by an arbitrary linear transformation. In fact, I leave it as a challenge (it won't be that hard) to find specific $c, c'$, choice of basis, and $f$ where the naturality condition fails.
Now recall the identity natural transformation. This is indeed a natural transformation since the choice of $\alpha_c=1_c$ makes no reference choice of basis. So we do not have to worry about odd choices being preserved.
For another example, see Emily Rhiel's "Category Theory in Context". This book is free on here website, so I absolutely recommend getting it and taking a peak at it since it does present things differently that MacLane. The section on naturality has a good example.
Basically, this commutativity condition requires that any choice we make in defining $\alpha_c$ and $\alpha_{c'}$ is preserved by $Ff$ and $Gf$, for every $f : c \to c'$.
I should add, though, that natural transformations can take on a very abstract and more general nature than the examples that motive the "freedom form choices" intuition. E.g., in posets or in more abstract categories. In these cases, it is sometimes hard to keep the original intuition. So, the naturality condition does not in general express the idea of "freedom form choices", but rather it puts a coherence condition on the assignment of objects to morphisms. This is similar to how a functor has coherence conditions on the assignment of morphisms to morphisms (the composition laws). So the full answer to your question is this: the naturality condition does not in and of itself assert a freedom of choices condition. This only makes sense in certain concrete categories. The naturality condition just requires a coherence between the choice of morphism. That this "coherence" idea is the proper way to think of the naturality condition is justified by its fitting into a much larger hierarchy. For more, see this and this.
Best Answer
You seem to have two questions. The first seems to be the following:
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)
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.