Let $\mathcal T$ and $\mathcal T'$ be triangulated categories and consider an additive functor $F:\mathcal T \to \mathcal T'$. Some authors say $F$ is triangulated if there exists a natural isomorphism $F(-)[1] \cong F(-[1])$ with some compatibility with distinguished triangle, i.e., it is a property of a functor while some say a triangulated functor is a pair of $F$ and such a natural isomorphism, i.e., it is an additional structure.
So, my questions is (please see edit) if these definitions are equivalent in the sense that there is a unique choice of such a natural isomorphism up to unique natural isomorphism. I think if $\mathcal T$, $\mathcal T'$, and $F$ have dg-enhancements, then the answer is yes since then such a natural isomorphism should be induced by the universal property of cartesian square, but I would really appreciate it if you would kindly correct me if this argument is not working.
Thank you in advance.
Edit: I think the first attempt of my question was not formulated as I wanted. Let me give a precise question.
Definition: Suppose we have additive functors $F,G:\mathcal T\to \mathcal T'$ of triangulated categories and natural isomorphisms $\epsilon_F:F(-[1])\cong F(-)[1]$ and $\epsilon_G:G(-[1])\cong G(-)[1]$. We say $(F,\epsilon_F)$ and $(G,\epsilon_G)$ are triangulated-naturally isomorphic if there is a natural isomorphism $\alpha:F\cong G$ such that for any $M \in \mathcal T$, the following diagram commute: $$\require{AMScd} \begin{CD} F(M[1]) @>{\epsilon_{F,M}}>> F(M)[1]\\ @V{\alpha_{M[1]}}VV @V{\alpha_M[1]}VV \\ G(M[1]) @>{\epsilon_{G,M}}>> G(M)[1]. \end{CD}$$
Question: Does the triangulated-natural isomorphism class of $(F,\epsilon_F)$ depend on the choice of $\epsilon_F$?