@SaalHardali IIRC in a stable setting the equation $Hom(X,Z)=DX⊗Z$ is true for all X iff Z is dualizable (this is of course different than the unstable setting, like R-modules!)
This was worked out in some MO answer, but I cannot remember which one...
@SaalHardali So, let's see if I can whip up a proof. If $Z$ is dualizable we have $[T,DX⊗Z]=[T⊗DZ⊗X,1]=[T⊗X,Z]=[T,F(X,Z)]$ for every $T$, so $F(X,Z)=DX⊗Z$
The viceversa should follow from taking the unit map $1→F(X,X)=DX⊗X$ and show it satisfies the triangular identities
@DenisNardin I'm cometely convinced. So much so that my initial example is in fact wrong because the same proof works in any symmetric monoidal category. A counter example to my assertion is Hom(Q/Z,Q/Z) \tensor Q != Hom( Q/Z, Q/Z \tensor Q)
Also I have a terrible Déjà Vu of asking the exact same question and coming to this exact realization before.