I know of several results to the effect that two triangulated categories are equivalent categories (usually one coming from algebra and one coming from topology). However, it's never been clear to me how to classify the possible different triangulated structures lifting a given "graded" category (a category enriched in graded abelian groups).
For example, how many different triangulated structures can there be with underlying category the derived category of Z-modules? All objects in this category are equivalent to direct sums of Z-modules concentrated in different degrees.
As another example, we have the category of "Z/2-graded abelian groups", which is equivalent to the derived category of differential graded modules over a Laurent polynomial ring on a generator in degree 2. There's another, equivalent-looking, homotopy category of modules over the periodic complex K-theory spectrum KU. These have equivalent underlying categories but I have never been clear on whether the triangulated structures are the same.
Best Answer
Generally speaking a unique lifting does not exist and I believe it is open as to what the possible liftings can be.
As an example of the non-uniqueness consider a slight variant of the particular category you asked about - namely $K^b(\mathbb{Z})$ the homotopy category of bounded complexes of finitely generated abelian groups. Considering this as a graded category it has at least 2 different structures of triangulated category. The usual one and one we denote $K^b(\mathbb{Z})^{-}$ where we declare that
$$X \stackrel{u}{\to} V \stackrel{v}{\to} Y \stackrel{w}{\to} \Sigma X$$ is a triangle if and only if
$$X \stackrel{-u}{\to} V \stackrel{-v}{\to} Y \stackrel{-w}{\to} \Sigma X$$ is a triangle with respect to the usual triangulation. The point that was implicit (before this edit) is that these triangulations do not agree. This also works for $D^b(\mathbb{Z})$, the bounded derived category of finitely generated abelian groups, indeed one just needs to suppose the two categories agreed and consider [TR3] applied to a diagram obtained by mapping $$\mathbb{Z} \stackrel{3}{\to} \mathbb{Z} \to K(3) \to \Sigma \mathbb{Z}$$ to $$\mathbb{Z} \stackrel{-3}{\to} \mathbb{Z} \to K(3) \to \Sigma \mathbb{Z}$$ by $(1,-1,h,1)$ where $h$ is some map provided by [TR3] but one can check this is ridiculous.
If you want more lifts of the suspended category with the usual suspension to a triangulated category I feel like the answer should be there aren't any but I don't know a proof or where this is written down if it is true. I think I know how to make some headway on the problem more generally if one rigidifies the situation a little but this is still work in progress.
More generally one can play this game with any unit in the degree zero part of the central ring of our suspended category (in plainer language we can twist by automorphisms of the identity functor commuting with suspension). In this note Balmer shows that this leads to suspended categories with infinitely many triangulations. I believe that it is unknown if there can be other sorts of triangulations at least on an indecomposable category.
I currently can't think of any examples that arise naturally and I am not familiar enough with the one you mention to be able to say at the moment whether or not the triangulations agree. If the suspensions agree then I'd guess they are equivalent up to something of the form above. I'd be interested to know if they weren't.
Update Here is the sort of thing I now have in mind for the bounded derived category of finitely generated abelian groups. As a warning I haven't really thought this through too thoroughly.
The statement I think one should be after is that no matter the triangulation there is no choice involved in the isomorphism class of the cones - more precisely if one is given $f\colon X\to Y$ then $\mathrm{cone}(f)$ is up to isomorphism independent of the triangulation. It should follow from this that the sign trick that gives you the second triangulation is all the wiggle room one has.
Denote by $\#X$ the non-negative integer $\lvert \{i\in \mathbb{Z} \;\vert\; H^i(X)\neq 0\} \rvert$, and try to run an induction on this quantity. If $X$ is the suspension of a group and we are given $f\colon X\to Y$ we can desuspend and assume it is in degree zero since this will just rotate the triangle. Complete $f$ to a triangle $$ X \stackrel{f}{\to} Y \to Z \to \Sigma X$$ Applying $\mathrm{Hom}(\mathbb{Z},-)$ gives an exact sequence of abelian groups which tells us that (where since we are not assuming the standard triangulation the use of cohomological notation is slightly abusive but just denotes the corresponding graded piece) $H^i(Z) \cong H^i(Y)$ for $i\leq -2$ and $i\geq 1$, that $H^0(Z) \cong \mathrm{coker} \; H^0(f)$, and that $H^{-1}(Z)$ is determined up to extension by $$0 \to H^{-1}(Y) \to H^{-1}(Z) \to \mathrm{ker}\;H^0(f) \to 0$$ but there is still no choice involved here since the element of $\mathrm{Ext}^1$ giving this extension is determined by the "degree 1 part" (what I mean by this is if $X$ has torsion it might map up to the piece of $Y$ in degree 1) of $f$ (I think - this is the bit I didn't really check). Thus the graded pieces of $Z$ are determined up to isomorphism and hence so is $Z$ since everything is a sum of its cohomology.
Now if we assume we have uniqueness for maps out of guys with less than $n$ pieces and let $\#X = n$ and suppose we are given $f\colon X\to Y$ which completes to a triangle with cone $Z$. Write $X$ as $X' \oplus X''$ where $\#X' < n$. Using the octahedral axiom we get a triangle (we can actually cook up 4 in this way but we only need 1) $$W \to \Sigma^{-1}Z \to X' \to \Sigma W$$ and since $\#X' < n$ the inductive hypothesis implies $Z$ is uniquely determined up to isomorphism.
I'd be interested if anyone has any ideas for an invariant that would detect the difference between these two triangulations on $D^b(\mathbb{Z})$ - they are really almost the same in every way I can think of except that the mapping axiom can fail between the triangles in each of them.