Let $T$ be a triangulated category admitting arbitrary coproducts with a set of compact generators $\mathcal{G}$. From the definition of compact generators we have that for any object $X$ of $T$ $[\Sigma^nZ,X]=0$ for all $n \in \mathbb{Z}$ and all $Z \in \mathcal{G}$ then $X=0$, equivalently $\mathcal{G}$ detects isomorphisms i.e. if a map $f \colon X \rightarrow Y$ has the property that $[\Sigma^nZ,f]$ is an isomorphism $\forall n \in \mathbb{Z}$ and $\forall \in \mathcal{G}$ then $f$ is an isomorphism.
But what if we have $[\Sigma^nZ,f]=0$? Can we deduce that $f$ is the zero map? I never found this claim anywhere and I could not prove it. I tried to show that the full subcategory of $T$ generated by $\{ A \in T : [A,f]=0 \}$ is a localizing subcategory but I have difficulties proving that is triangulated. Basically I get in a situation similar to the 5 lemma but where the two vertical maps on the left and on the right are zero and I cannot conclude the central one is zero from the exactness of the long horizontal sequences.
Thus I am induced to think that the claim is false: can you provide an explicit counterexample?
Also what if I add more structures or properties to $T$? For example I could consider a closed tensor triangulated category such that the elements of $\mathcal{G}$ are strongly dualizable or such that every cohomology functor is representable.
If you allow yourself to choose the set $\mathcal{G}$ of compact generators then there are very simple examples. For example, take $T=D(\text{Mod }R)$ for some ring $R$, and $\mathcal{G}=\{\Sigma^nR\mid n\in\mathbb{Z}\}$, and let $f:M\to\Sigma N$ be a map representing a nonzero element of $\text{Ext}^1_R(M,N)$.
But even if you take $\mathcal{G}$ to be the class of all compact objects, then there are examples even in quite normal triangulated categories, that have been quite extensively studied under the name "phantom maps".