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This question was inspired by this question. The question whether the connecting morphism in the snake lemma is unique needs to be formulated precisely. Of course, in any specific exact sequence coming from the snake lemma, there may be a whole lot of other morphisms making the sequence exact, but this isn't very satisfying. Hagen von Eitzen gives the much more interesting example $-\delta$ in the linked question, which happens to be natural as well. My question essentially is how unique the connecting morphism from the snake lemma becomes when imposing this naturality condition.

To set the stage, fix some abelian category. Consider the category of diagrams with exact rows in this abelian category of the form $$\require{AMScd} \begin{CD} @. A @>{f}>> B @>{g}>> C @>>> 0\\ @. @VV{a}V @VV{b}V @VV{c}V \\ 0@>>> A' @>>{f'}> B' @>>{g'}> C' \end{CD}$$ The morphisms in this category are morphisms of diagrams. The snake lemma construction in fact gives us a functor from this category to the category of six-term exact sequences in the abelian category (morphisms again being morphisms of diagrams). Explicitly, it maps such a diagram to the exact sequence $$\ker a\rightarrow\ker b\rightarrow\ker c\stackrel{\delta}{\rightarrow}\operatorname{coker}a\rightarrow\operatorname{coker}b\rightarrow\operatorname{coker}c.$$ The unlabeled morphisms are the ones induced by the morphisms in the diagrams. Similarly, a morphism between these diagrams induces a morphism between the associated six-term sequences whose components are induced by the components of the original morphism. Now we can make the question precise: how many such functors are there acting the same way on objects and morphisms except for possibly associating a different connecting morphism $\delta^{\prime}\colon\ker c\rightarrow\operatorname{coker}a$ to each diagram?

As a preliminary observation, note that if $\psi$ is an automorphism of the identity functor of the abelian category, i.e. a choice of automorphism $\psi_A\colon A\rightarrow A$ for every object $A$ such that $\psi_B\circ f=f\circ\psi_A$ for any morphism $f\colon A\rightarrow B$, we can consequently replace $\delta$ by $\psi_{\operatorname{coker}a}\circ\delta=\delta\circ\psi_{\ker c}$ and obtain another such functor. In the category of abelian groups, additive inversion provides such an automorphism and this recovers Hagen von Eitzens example mentioned earlier.

Conversely, if we have such a functor, then to the diagram $$\require{AMScd} \begin{CD} @. 0 @>>> A @= A @>>> 0\\ @. @VVV @| @VVV \\ 0@>>> A @= A @>>> 0 \end{CD}.$$ there is the associated exact sequence $$0\rightarrow0\rightarrow A\stackrel{\psi_A}{\rightarrow}A\rightarrow0\rightarrow0.$$ Functoriality implies naturality of the map $\psi_A$ and it is an automorphism by exactness. Thus, we see that each such functor induces an automorphism of the identity functor.

If we start with an automorphism of the identity functor, pass to the functor and then pass back to an automorphism of the identity functor via these constructions, we end up where we started again. If we carry out the two constructions in the reverse order, it is not as clear what happens. That is, is such a functor determined by how it acts on diagrams of the specific shape from the last paragraph? Are there any such functors not arising from the construction described two paragraphs ago? And, ultimately, how many such functors are there/can we give a meaningful characterization of them?

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This is a partial answer, whose contents seem fairly obvious in hindsight, but since they didn't occur to me at the time of posting the question, I deem them worth recording. In the following, $\delta$ always means the "usual" connecting morphism (and no attempt at distinguishing the connecting morphisms in different diagrams or even different categories will be made notationally). Firstly, the question has a positive answer in the category $\mathbf{Mod}(R)$ of $R$-modules for a given ring $R$.

Theorem: If $R$ is a ring, the connecting homomorphisms in $\mathbf{Mod}(R)$ are precisely $r\delta$, where $r\in R^{\times}$.

Proof: It is clear that these are connecting homomorphisms as multiplication by a unit is an automorphism of the identity functor. Conversely, let $\delta^{\prime}$ be any connecting homomorphism. First, the diagram \begin{CD} @. 0 @>>> R @= R @>>> 0\\ @. @VVV @| @VVV \\ 0@>>> R @= R @>>> 0 \end{CD} induces an exact sequence \begin{CD} 0 @>>> 0 @>>> R @>{\delta^{\prime}}>> R @>>> 0 @>>> 0. \end{CD} By exactness, $\delta^{\prime}\colon R\rightarrow R$ is an isomorphism, hence given as multiplication by a unit $r\in R^{\times}$. Next, consider an arbitrary diagram \begin{CD} @. A @>{f}>> B @>{g}>> C @>>> 0\\ @. @VV{a}V @VV{b}V @VV{c}V \\ 0@>>> A' @>>{f'}> B' @>>{g'}> C' \end{CD} with exact rows. Let $z\in\ker(c)$ be arbitrary. Since $g\colon B\rightarrow C$ is surjective, there exists a $y\in B$ such that $g(y)=z$. Next, $g^{\prime}(b(y))=c(g(y))=c(z)=0$, so, by exactness, there exists an $x^{\prime}\in A^{\prime}$ such that $f^{\prime}(x^{\prime})=b(y)$. Since $R$ is a free $R$-module on one generator, we obtain homomorphisms $R\rightarrow C,\,1\mapsto z$, $R\rightarrow B,\,1\mapsto y$, $R\rightarrow B^{\prime},\,1\mapsto b(y)$ and $R\rightarrow A^{\prime},\,1\mapsto x^{\prime}$. Now, by construction, these define a morphism of diagrams from the first to the second of the just discussed diagrams, so induce the following homomorphism of exact sequences \begin{CD} 0 @>>> 0 @>>> R @>{r\cdot}>> R @>>> 0 @>>> 0\\ @VVV @VVV @VVV @VVV @VVV @VVV\\ \ker(a) @>>> \ker(b) @>>> \ker(c) @>{\delta^{\prime}}>> \mathrm{coker}(a) @>>> \mathrm{coker}(b) @>>> \mathrm{coker}(c). \end{CD} The middle square now yields $1\mapsto r\mapsto r[x^{\prime}]$ one way round and $1\mapsto z\mapsto\delta^{\prime}(z)$ the other way round. Since, by construction, $[x^{\prime}]=\delta(z)$, and the diagram was arbitrary, we obtain $\delta^{\prime}=r\delta$.

Secondly, we often want to work in multiple abelian categories at once. Thus, it might seem desirable to ask for a choice of connecting morphism simultaneously in all abelian categories that is compatible with exact functors (in the sense that exact functors preserve the connecting morphisms). These can be classified too.

Theorem: The only simultaneous choices of connecting morphism in all abelian categories that are compatible with exact functors are $\delta$ and $-\delta$.

Proof: That these two choices work is standard. Conversely, assume such choices $\delta^{\prime}$ have been made. For any ring $R$, the forgetful functor $\mathbf{Mod}(R)\rightarrow\mathbf{Mod}(\mathbb{Z})$ is exact. Since the only connecting homomorphisms in $\mathbf{Mod}(\mathbb{Z})$ are $\pm\delta$ by the previous Theorem, this also has to be the choice in any of the categories $\mathbf{Mod}(R)$. Let $\mathcal{A}$ be an abelian category and consider an arbitrary diagram of the usual shape in $\mathcal{A}$. Then we can choose a small, abelian subcategory with exact inclusion $\mathcal{B}\rightarrow\mathcal{A}$ and such that the diagram lives in $\mathcal{B}$ (see this question). By the Freyd-Mitchell embedding theorem, there is an exact embedding $F\colon\mathcal{B}\rightarrow\mathbf{Mod}(R)$ for some ring $R$. Since the connecting morphism in $\mathbf{Mod}(R)$ is $\pm\delta$ and $F$ is additive and preserves $\delta$, we have $F(\delta^{\prime})=\pm\delta=F(\pm\delta)$. Since $F$ is faithful, we have $\delta^{\prime}=\pm\delta$ and since the choices were arbitrary, the claim follows.

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