$\require{AMScd}$
Let $\mathcal{T}$ be a triangulated category, and $\mathcal{S}$ a full subcategory of $\mathcal{T}$ (which is not triangulated).
Let $F, G: \mathcal{T} \to \mathcal{T}$ be two triangulated functors.
I have a "partial" natural isomorphism: For any $M \in \mathcal{S}$, I am given an isomorphism $\eta_M: F(M) \to G(M)$, such that if $f:M \to N$ is a morphism in $\mathcal{S}$, the diagram $$ \begin{CD} F(M) @>{F(f)}>> F(N)\\ @VVV @VVV \\ G(M) @>{G(f)}>> G(N) \end{CD} $$ is commutative.
Now, I am given two distinguished triangles $$\mathbf{M} = M' \to M'' \to M \to \Sigma M'$$ and $$\mathbf{N} = N' \to N'' \to N \to \Sigma N'$$ with $M',M'',N',N'' \in \mathcal{S}$, but $M,N \notin \mathcal{S}$.
I am also given a morphism of triangles $\mathbf{f} = (f',f'',f): \mathbf{M} \to \mathbf{N}$.
Using the triangles $F(\mathbf{M}), G(\mathbf{M})$, and the isomorphisms $\eta_{M'}, \eta_{M''}$, I can obtain an isomorphism $\eta_M: F(M) \to G(M)$, and similarly an isomorphism $\eta_N:F(N) \to G(N)$.
My question:
Is it possible to choose the isomorphisms $\eta_M, \eta_N$ so that $G(f) \circ \eta_M = \eta_N \circ F(f)$?
