Extending natural transformations of triangulated functors on $D^b(\mathbb P^1)$

Question

Let $\mathcal T = D^b(\mathbb P^1)$, the bounded derived category of $\mathbb P^1$ over a field $k$. It is well known that $\mathcal T$ admits a strong and full exceptional sequence $\{\mathcal O, \mathcal O(1) \}$, such that $\mathrm{Hom}(\mathcal O, \mathcal O(1))$ is 2-dimensional over $k$. Now, let $\mathcal T'$ be another triangulated category, and let $F, G : \mathcal T \to \mathcal T'$ be exact functors. Write $i : \{ \mathcal O, \mathcal O(1) \} \hookrightarrow \mathcal T$ for the inclusion functor. At least under some reasonable assumptions, I can prove that a natural tranformation $\varphi : F \circ i \to G \circ i$ can be extended to a natural transformation $F \to G$ (this is true when $F$ and $G$ are enhanceable to dg-functors, and the proof I know involves those enhancements); what if I try to extend a natural transformation defined on the full subcategory $\{ \mathcal O, \mathcal O(1), C(x_0) \}$, where $C(x_0)$ is the cone of the morphism $x_0 : \mathcal O \to \mathcal O(1)$? Formally: let $j : \{ \mathcal O, \mathcal O(1), C(x_0) \} \hookrightarrow \mathcal T$ be the inclusion, and let $\varphi : F \circ j \to G \circ j$ be a natural transformation. Does an extension $\widetilde{\varphi} : F \to G$ of $\varphi$ (such that $\widetilde{\varphi} \circ j = \varphi$) exist? How can I construct it? If necessary, I can also assume that $F$ and $G$ come from dg-functors between suitable enhancements.