Question on the fill-in morphism in a triangulated category

Let
$$
\begin{array}{rcl}
A&\to& B\\
\downarrow & &\downarrow\\
A’&\to& B’
\end{array}
$$
be a commutative diagram in a triangulated category. By the axioms of a triangulated category, this may be ‘completed’ to
$$
\begin{array}{rccccl}
A&\to& B&\to& C&\to\\
\downarrow & &\downarrow&&\downarrow\\
A’&\to& B’&\to& C’&\to
\end{array}
$$
where both rows are exact triangles. Let
$$
\begin{array}{rccccl}
A&\to& B&\to& \tilde C&\to\\
\downarrow & &\downarrow&&\downarrow\\
A’&\to& B’&\to& \tilde C’&\to
\end{array}
$$
be another ‘completion’ of the initial diagram such that both rows are exact triangles. One gets that $C\cong \tilde C$ and $C’\cong \tilde C’$. My question is: May these two isomorphism be chosen in a way that the diagram
$$
\begin{array}{rcl}
C&\overset{\cong}{\to}& \tilde C\\
\downarrow & &\downarrow\\
C’&\overset{\cong}{\to}& \tilde C’
\end{array}
$$
commutes?

Solutions Collecting From Web of "Question on the fill-in morphism in a triangulated category"

Not in general.

For example, let the original commutative square be
$$\begin{array}{ccc}
X[-1]&\to&0\\
\downarrow & &\downarrow\\
0&\to&X
\end{array}$$
for any non-zero object $X$.

This can be completed to a diagram
$$\begin{array}{cccccccc}
X[-1]&\to& 0&\to& X&\to &X\\
\downarrow & &\downarrow&&\downarrow&&\downarrow\\
0&\to& X&\to& X&\to& 0
\end{array}$$
with a completely free choice of the vertical map $X\to X$.

Taking two different completions, where this map is zero in one and the identity in the other, gives an example where the answer to your question is “no”.