@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!)

@DenisNardin I too vaguely recall something of the sort, but didn't find it either and then I became skeptical.

@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.

Here's simple proof for me to never forget this:

$Hom(X,Z)=Hom(X \otimes D(Z), 1)=Hom(D(Z),D(X))=D^2(Z) \otimes D(X)= Z \otimes D(X) = D(X) \otimes Z$

(What I meant is to write a mnemonic for your proof Denis of course)