Let $\mathcal C,\otimes$ be a monoidal category, i.e. $\otimes : \mathcal C \times \mathcal C \to \mathcal C$ is a functor, and there's a bit more structure and properties. Suppose that for each $X \in \mathcal C$, the functor $X \otimes - : \mathcal C \to \mathcal C$ has a right adjoint. I will call this adjoint (unique up to canonical isomorphism of functors) $\underline{\rm Hom}(X,-) : \mathcal C \to \mathcal C$. By general abstract nonsense, $\underline{\rm Hom}(X,-)$ is contravariant in $X$, and so defines a functor $\underline{\rm Hom}: \mathcal C^{\rm op} \times \mathcal C \to \mathcal C$. If $1 \in \mathcal C$ is the monoidal unit, then $\underline{\rm Hom}(1,-)$ is (naturally isomorphic to) the identity functor.

Then there are canonically defined "evaluation" and "internal composition" maps, both of which I will denote by $\bullet$. Indeed, we define "evaluation" $\bullet_{X,Y}: X\otimes \underline{\rm Hom}(X,Y) \to Y$ to be the map that corresponds to ${\rm id}: \underline{\rm Hom}(X,Y) \to \underline{\rm Hom}(X,Y)$ under the adjuntion. Then we define "composition" $\bullet_{X,Y,Z}: \underline{\rm Hom}(X,Y) \otimes \underline{\rm Hom}(Y,Z) \to \underline{\rm Hom}(X,Z)$ to be the map that corresponds under the adjunction to $\bullet_{Y,Z} \circ (\bullet_{X,Y} \otimes {\rm id}) : X \otimes \underline{\rm Hom}(X,Y) \otimes \underline{\rm Hom}(Y,Z) \to Z$. (I have supressed all associators.)

Question:Is $\bullet$ an associative multiplication? I.e. do we have necessarily equality of morphisms $\bullet_{W,Y,Z} \circ (\bullet_{W,X,Y} \otimes {\rm id}) \overset ? = \bullet_{W,X,Z} \circ ({\rm id}\otimes \bullet_{X,Y,Z})$ of maps $\underline{\rm Hom}(W,X) \otimes \underline{\rm Hom}(X,Y) \otimes \underline{\rm Hom}(Y,Z) \to \underline{\rm Hom}(X,Z)$? If not, what extra conditions on $\otimes$ are necessary/sufficient?