Let $\mathcal{C}$ and $\mathcal{D}$ be triangulated categories, and suppose that there exists an additive equivalence $F: \mathcal{C} \to \mathcal{D}$. Suppose further that $\mathcal{C} = \text{add}(C_1 \oplus \dots \oplus C_n)$ and $\mathcal{D} = \text{add}(D_1 \oplus \dots \oplus D_n)$, where the $C_i$ and $D_i$ are indecomposable, where here $\text{add}$ denotes additive closure. Finally, suppose that the translation functors on $\mathcal{C}$ and $\mathcal{D}$ permute the $C_i$ and $D_i$ respectively, that $F(C_i) = D_i$ for all $i$, and that $F(C_i[1]) = D_i[1]$ for all $i$.

Does it follow that $F$ is in fact a triangulated equivalence? If not, are there any reasonable hypotheses one can place on $\mathcal{C}$ and $\mathcal{D}$ to ensure that it is?

Let $\mathcal{C}$ and $\mathcal{D}$ be triangulated categories, and suppose that there exists an additive equivalence $F: \mathcal{C} \to \mathcal{D}$. Suppose further that $\mathcal{C} = \text{add}(C_1 \oplus \dots \oplus C_n)$ and $\mathcal{D} = \text{add}(D_1 \oplus \dots \oplus D_n)$, where the $C_i$ and $D_i$ are indecomposable, where here $\text{add}$ denotes additive closure. Finally, suppose that the translation functors on $\mathcal{C}$ and $\mathcal{D}$ permute the $C_i$ and $D_i$ respectively, that $F(C_i) = D_i$ for all $i$, and that $F(C_i[1]) = D_i[1]$ for all $i$.

Does it follow that $F$ is in fact a triangulated equivalence? If not, are there any reasonable hypotheses one can place on $\mathcal{C}$ and $\mathcal{D}$ to ensure that it is?

Edit: A related question is the following. Suppose that $\mathcal{C} = \text{add}(C_1 \oplus \dots \oplus C_n)$ is a triangulated category admitting two triangulated structures, with shifts $[1]$ and $[1]'$, and where the $C_i$ are indecomposable. Suppose that these shifts permute the $C_i$ and that $C_i [1] = C_i[1]'$ for all $i$. Does it follow that the triangulated structures coincide?