Does a functor which is additive by arbitrary isomorphism restrict and corestrict to an actually additive functor? (I appologize for asking a more and more refined version of the same problem for the third time in two days. This is just due to the fact that with every answer to a previous question, I realize that the question I asked had some sort of loophole, which allowed for a construction that, although very interesting, wasn't exactly what I'm looking for. I'm trying my best to ask the question I intended to ask now, but let me know if you consider my repeated posting of very similar but slightly different questions inappropriate. Now to the question.)
Suppose $F$ is a functor from the category of abelian groups to itself, such that


*

*For all abelian groups $A$ and $B$, we have $F(A\oplus B)\cong F(A)\oplus F(B)$ by an arbitrary isomorphism 

*$F(\mathbb{Z})$ is finitely generated and free.


Now let $\mathcal{F}$ be the full subcategory of the category of abelian groups, whose objects are the finitely generated, free abelian groups. The two properties imply that $F$ restricts and corestircts to a functor $F:\mathcal{F}\to\mathcal{F}$, which by a slight abuse of notation I still denote by $F$. Is it true that $F$ is additive, in the sense that for all split exact sequences
$$
0\to A\overset{i}{\to} A\oplus B\overset{p}{\to} B\to 0
$$
with $A,B\in\operatorname{Ob}(\mathcal{F})$ and $i:a\mapsto(a,0)$ and $p:(a,b)\mapsto b$, is the sequence
$$
0\to F(A)\overset{Fi}{\to} F(A\oplus B)\overset{Fp}{\to} F(B)\to 0
$$
 still split exact?
Note that both the construction in my first question as well as the very clever construction of Eric Wofsey in my second question don't provide a counterexample.
I guess one thing I am trying to figure out in this sequence of questions, and for which I don't have yet a good intuition about, is how "rigid" the notion of a functor is. More precisely, when first tackling the problem in my first question, I had a very hard time comming up with a functor which is nice enough to preserve finite direct sums by an arbitrary isomorphism, but pathological enough to do this not via the induced maps of the original split exact sequence. So this question is pushing this to the limit, because $\mathcal{F}$ is so small that $F$ is essentially a function, but still I see no way of proving that split exact sequences are preserved.
 A: Yes.
First, it follows from the conditions that $F$ preserves being zero (for objects and maps) since $F(0)\cong F(0)\oplus F(0)$ and $F(0)$ is a finitely generated free abelian group.
So $(Fp)(Fi)=0$. Let $\alpha:A\oplus B\to A$ be projection onto $A$ and $\beta:B\to A\oplus B$ the inclusion of $B$, so that $\alpha i=\text{id}_A$, $p\beta=\text{id}_B$, $pi=0$ and $\alpha\beta=0$, and since $F$ preserves identy maps and zero maps, we get the same equations after applying $F$ to $p$,$i$,$\alpha$ and $\beta$.
Since $\varphi:F(A\oplus B)\to\text{im}(Fi)\oplus\text{im}(F\beta)$, given by $\varphi(x)=(FiF\alpha(x),F\beta Fp(x))$, restricts to the identity on $\text{im}(Fi)\oplus\text{im}(F\beta)$, which is isomorphic to $F(A)\oplus F(B)$, we deduce that $\text{im}(Fi)\oplus\text{im}(F\beta)$ is a summand of $F(A\oplus B)$, which is also isomorphic to $F(A)\oplus F(B)$. 
Since everything is a finitely generated free abelian group, $\text{im}(Fi)\oplus\text{im}(F\beta)=F(A\oplus B)$, and so
$$0\to F(A)\overset{Fi}{\to} F(A\oplus B)\overset{Fp}{\to} F(B)\to 0$$
is split exact.
