Let $Z$ be an object in a stable (or triangulated/whatever) category $\mathcal C$. I believe it follows from Thomason's theorem that the triangulated categories generated by $$Y = Z \oplus \Sigma Z$$ and $$X = Y \oplus \Sigma Y = Z \oplus \Sigma Z \oplus \Sigma Z \oplus \Sigma^2 Z$$ are the same. This means that $X$ and $Y$ can each be constructed from one one another in finitely many steps, where each step consists of taking the fiber or cofiber of a map between previously-constructed objects. One direction of this is obvious -- clearly $X$ can be constructed from $Y$ in this way. The other direction is not so clear to me.

Question: Let $Z$ be an object in a stable category $\mathcal C$, and let $Y,X$ be as above. How can one explicitly construct $Y$ from $X$ in finitely many steps, using just fibers and cofibers?

(It takes finitely many steps to construct a de/suspension, extension, or direct sum from fibers and cofibers, so one is free to use these as steps as well.)