First of all, let me fix some notation.

Let $\mathcal D$ be a triangulated category in the sense of Verdier-Grothendieck (for example, the homotopy category $\mathbf{K}(k)$ of cochain complexes over a fixed commutative ring $k$). I call

*cone*of a morphism $f : A \rightarrow B$ the object $C(f)$ (uniquely determined up to isomorphism) such that $A \stackrel{f}\rightarrow B \rightarrow C(f) \rightarrow A[1]$ is a distinguished triangle in $\mathcal D$. When $\mathcal D = \mathbf{K}(k)$, $C(f)$ can be identified (up to homotopy equivalence) to the mapping cone of the chain map $f$.In any category $\mathcal C$, I say that two morphisms $f: A \to B$ and $f' : A' \to B'$ are isomorphic, if they are isomorphic in the category of morphisms $\mathrm{Mor}(\mathcal C)$, that is, there are two isomorphisms $u : A \to A'$ and $v : B \to B'$ such that $vf = f'u$.

Now, my question is the following: is there a triangulated category with a pair of *parallel* morphisms $f,f' : A \to B$ such that $f$ is *not* isomorphic to $f'$ but $C(f)$ is isomorphic to $C(f')$? I believe that an example could be found in the category $\mathbf{K}(k)$.

Of course, if we don't require $f$ and $f'$ to be parallel, then we may find examples in any reasonable triangulated category: just set $f=1_0$, the identity of a zero object, and $f' = 1_A$, the identity of a nonzero object. Then, both cones are zero objects (a general fact in triangulated categories), but clearly $f$ is not isomorphic to $f'$.