If I have two exact triangles $X \to Y \to Z \to X[1]$ and $X' \to Y' \to Z' \to X'[1]$ in a triangulated category, and I have morphisms $X \to X'$, $Y \to Y'$ which 'commute' (i.e., such that $X \to Y \to Y' = X \to X' \to Y'$), thene there exists a (not necessarily unique) map $Z \to Z'$ which completes what we've got to a morphism of triangles.
Is there a criterion which ensures the uniqueness of this cone-map?
I'd like something along the lines of: if $\operatorname{Ext}^{-1}(X,Y')=0$ then yes.
(I might be too optimistic, cfr. Prop 10.1.17 of Kashiwara-Schapira Categories and Sheaves: in addition to $\operatorname{Hom}{(X[1],Y')} = 0$ they also assume $\operatorname{Hom} {(Y,X')} =0$. I really don't have this second assumption.)
(In the case I'm interested in $X=X', Y=Y'$ and $X\to X'$, $Y \to Y'$ are the identity maps.)
(If it makes things easier, although I doubt it, you can take the category to be the bounded derived category of coherent sheaves on some, fairly nasty, scheme.)
In the context I have in mind $X, Y, X', Y'$ are all objects of the heart of a bounded t-structure. If we assumed $\operatorname{Hom}{(Z,Y')} = 0$ or $\operatorname{Hom}{(X[1],Z')} = 0$ then the result easily follows. I don't think I'm happy making those assumptions though.