When is a triangulated category uniquely determined by its generators? Notation: Let $\mathcal T$ be a triangulated category, and let $\mathcal E$ be a full subcategory of $\mathcal T$. I write $\langle \mathcal E \rangle$ to indicate the smallest strictly full triangulated subcategory of $\mathcal T$ which contains $\mathcal E$.
Now, assume we are given triangulated categories $\mathcal T= \langle \mathcal E \rangle$ and $\mathcal T' = \langle \mathcal E' \rangle$, such that the "generating subcategories" $\mathcal E$ and $\mathcal E'$ are equivalent (or even isomorphic). Is it true that $\mathcal T$ is equivalent to $\mathcal T'$? Or, if not in general, is it true under some reasonable assumptions?
 A: I don't think that there is any reasonable context in which one would expect this to be true.  For example, you can take $\mathcal{T}$ to be the category of spectra and $\mathcal{T}'$ to be the derived category of $\mathbb{Z}$, then put $\mathcal{E}=\{S^0\}$ and $\mathcal{E}'=\{\mathbb{Z}\}$.  Then $\mathcal{E}\simeq\mathcal{E}'$ but $\mathcal{T}\not\simeq\mathcal{T}'$.
If you want to insist that $\mathcal{E}$ is closed under (de)suspension, or regard $\mathcal{T}$ as a category enriched over graded abelian groups, then the above example breaks down, but there are other counterexamples.  I think you can get some by considering the stable module categories of nonisomorphic $p$-groups with isomorphic mod $p$ Tate cohomology, such as $C_{p^2}$ and $C_{p^3}$.
A: Maybe someone who knows more will come around and answer this in more detail, but I believe the short answer is:

No.  This is why triangulated categories belong in the ashbin of history, and we should have been using dg-categories all along.

The example you should have in mind is essentially the minimal one: consider the triangulated category of compact dg-modules over two dg-algebras $A$ and $B$ which are not quasi-isomorphic but have isomorphic cohomology.  These at least shouldn't be the same as triangulated categories (maybe someone can helpfully supply a proof), but $A$ and $B$ themselves are generating subcategories which are equivalent from the triangulated perspective, since the cohomology of $A$ and of $B$ are the relevant Ext-spaces.  
For dg-categories on the other hand, this sort of property is essentially built-in from the start.
