Let $D$ be a triangulated category where a triangle is by definition $X,Y,Z\in D$ and $T:D\to D$ automorphism s.t. $X\to Y\to Z\to T(X)$ is called a triangle.
Consider opposite functor(contravariant) $op:D\to D^{op}$.
Say $X\to Y\to Z\to T(X)$ is a distinguished triangle. Under $op$ map it is being mapped to $T(X)^{op}\to Z^{op}\to Y^{op}\to X^{op}$.
$\textbf{Q:}$ Why is the image of a distinguished triangle is distinguished here? Note that definition of triangulated category's morphism follows one direct. However, when one takes opposite functor image, all arrows are flipped. In particular, a triangle does not look like a triangle unless one insists definition of triangle in opposite category looks like $T(X)\to Z\to Y\to X$ instead of $X\to Y\to Z\to T(X)$. It should be clear that $T$ commutes with $op$.(i.e. We have two choice of automoprhisms on $D^{op}$ here. One is $T$ and the other one is $T^{-1}$.) Since $Id_X\cong TT^{-1}$. Apparent choice in $D^{op}$'s is $T^{-1}:D^{op}\to D^{op}$ as automorphism. However, due to $Id_X\cong TT^{-1}$ is a natural transformation, it is not at once clear that $T(X)\to Z\to Y\to X$ is isomorphic to $T(X)\to Z\to Y\to TT^{-1}(X)$. I guess invertible functor $T$ here really means $TT^{-1}=Id$.(Normally a shift functor has property that $[1]^2=Id$.)
Ref. pg 101 remark 6.1.5 of pdf. https://webusers.imj-prg.fr/~pierre.schapira/lectnotes/HomAl.pdf