Consider a map $f\colon S^1 \to S^1 \vee S^1$ that maps upper half of a circle to the first summand in orientation preserving way, and lower half to the second circle in orientation reversing way. Also denote $i_1$ and $i_2$ inclusions of $S^1$ to $S^1 \vee S^1$ as two summands. Composition of $f$ with folding map $p\colon S^1\vee S^1 \to S^1$ (both summands in orientation-preserving way) is evidently homotopic to a constant map. Looking on $H_1$ we have $\mathbb{Z} \xrightarrow{f_*} \mathbb{Z}\oplus \mathbb{Z} \xrightarrow{p_*} \mathbb{Z}$.
$p_*$ is addition: $S^1 \xrightarrow{i_1} S^1\vee S^1 \xrightarrow{p} S^1$ is identity (similary for $i_2$). $i_1$ induces $\mathbb{Z} \to \mathbb{Z}\oplus\mathbb{Z}$, to make sure it is an inclusion of the first summand we may use naturality of exact sequences for a map $(S^1, *) \xrightarrow{i_1} (S^1\vee S^1, \mathrm{im}\, i_2)$. Now we know $p_*(1, 0) = 1$ and $p_*(0, 1) = 1$, by linearity $p_*$ is addition.
Thus $f_*$ maps $1\in H_1(S^1)\cong \mathbb{Z}$ to some pair $(x, -x)\in H_1(S^1\vee S^1)$. To find out what $x$ is, consider the composition $S^1 \xrightarrow{f} S^1 \vee S^1 \to S^1$ where the second map collapses second summand to a point (it induces projection on a second coordinate in $\mathbb{Z}\oplus \mathbb{Z} \to \mathbb{Z}$ too, e.g. by naturality of exact sequences of pairs). By construction this map is homotopic to identity. Similary for the second summand in $S^1\vee S^1$ we get that orientation-reversing map induces $-1$ on homology.
That's quite cumbersome to read, but actually very straightforward.
UPDATE
As mentioned in comments, here is proof reversing orientation induces identity on $H_0$. $* \to S^1$ and $*\to S^1 \to S^1$ (the second map is reversal of orientation) are equal as maps from a point, thus induce the same map on $H_0$. If we know that the inclusion of a base point in $S^1$ induces non-zero map on $H_0$, we're done, since automorphism of $H_0(S^1) \cong \mathbb{Z}$ fixing a non-zero element must be identity. To see this, use the naturality of long exact sequences of pairs for a map $(S^1, *) \to (*, *)$. The first commutative square (with non-relative $H_0$s) consists of three identity mapping of a point and a map induced from inclusion of a base point, and cannot commute unless $H_0(*) \to H_0(S^1)$ is an inclusion.
(Reply to a comment: nullhomotopic map induces zero on $H_1$ since it factors as map $X \to * \to Y$)
Also it seems to me that $X \hookrightarrow X \vee Y$ induces inclusion to the first summand in homology not due to functoriality properties of long exact sequences but this is just the way we identify $H_i(X\vee Y)$ with $H_i(X)\oplus H_i(Y)$. That is, we look at the pair $(X \vee Y, Y)$, write long exact sequence, then apply excision to get rid of relative homology and get $H_i(Y)$ instead. After that, we note that a lot of arrows in long exact sequence admit splitting and define the isomorphism $H_i(X \vee Y) \to H_i(X) \oplus H_i(Y)$ the way inclusion of $X$ in a wedge sum is inclusion on first summand. Thus some arguments by naturality above are not needed.
The "natural" bit here is a red herring, and seeing that it's not relevant makes the question a lot easier.
Indeed at this level of generality ($\mathcal{L}$ an arbitrary category) we can let functors "be anything" and "natural" sort of loses its point. Indeed, it suffices to take $\mathcal{L}$ to be the discrete category on one object.
At that point, a functor $\mathcal{L}\to \mathsf{Comp}$ is just the choice of a complex, and $\mathcal{L}\to \mathbf{Ab}$ just the choice of an abelian group.
A natural isomorphism (or more generally transformation) is then just an isomorphism (or more generally morphism). Hence at this level of generality, a special case of your question becomes :
If I have an isomorphism between the homologies of two complexes, is this isomorphism induced by some (iso)morphism of complexes ?
The answer to that is clearly no. For instance consider a complex $\mathsf{C}$ with only $\mathbb{Z/2Z}$ in position $0$ and $0$'s elsewhere, and $\mathsf{C'}$with $\mathbb{Z}$ in position $-1$ and $0$, the map $\mathbb{Z}\to \mathbb{Z}$ being multiplication by $2$ . Then the homologies of both complexes are isomorphic (they're $\mathbb{Z/2Z}$ in position $0$, and $0$ elsewhere); but there is no nontrivial morphism $\mathsf{C\to C'}$, hence no morphism that could induce the isomorphism in homology.
In fact this isn't category-dependent (you might argue that I chose a dummy category): for any category $\mathcal{L}$, you can "model" this situation by picking constant functors, and the result will be the same; therefore if you want to add some conditions to change the answer to "yes", then the conditions will not only be on $\mathcal{L}$ but also on the functors you have, e.g. respect products or whatever. But that would be another question, and right now I don't have examples of natural (non trivial) constraints one could add on the functors to have a positive answer.
The moral here is that the functor $H_* : \mathsf{Comp}\to \mathbf{Ab}$ is very very far from being full.
Best Answer
As Tyrone remarked in his comment, we have the "restriction functor" $R : \operatorname{Pairs} \to \operatorname{Pairs}$ given by $R(X,A) = (A,\emptyset)$ and $R(f) )= f\mid_A : (A,\emptyset) \to (B,\emptyset)$ for $f : (X,A) \to (Y,B)$. Then $$\partial_n : h_n \to h_{n-1} \circ R$$ is a natural transformation between the functors $h_n : \operatorname{Pairs} \to \operatorname{Ab}$ and $h_{n-1} \circ R : \operatorname{Pairs} \to \operatorname{Ab}$.