The internal-hom composition morphism

Consider the composite $([Y,Z]\otimes [X,Y])\otimes X\cong [Y,Z]\otimes ([X,Y]\otimes X)\stackrel{1\otimes {\mathrm{ev}}_{X,Y}}{\to }[Y,Z]\otimes Y\stackrel{{\mathrm{ev}}_{Y,Z}}{\to }Z$. Transposing it across the adjunction $-\otimes X⊣[X,-]$ gives a unique morphism ${\mathrm{comp}}_{X,Y,Z}:[Y,Z]\otimes [X,Y]\to [X,Z]$.

Transpose the left unitor ${\lambda }_X:\mathbf{1}\otimes X\to X$ across the same adjunction to obtain $u_X:\mathbf{1}\to [X,X]$.

To compare the two composites $[Z,W]\otimes [Y,Z]\otimes [X,Y]\rightrightarrows [X,W]$, tensor each with $X$ and postcompose with ${\mathrm{ev}}_{X,W}$. Both have the same transpose, namely the triple evaluation map $[Z,W]\otimes [Y,Z]\otimes [X,Y]\otimes X\to W$, so the transposition bijection of [L1] forces the two composites to be equal.

The left and right unit laws are proved the same way: after tensoring with $X$ and evaluating, both candidate composites have transpose ${\mathrm{ev}}_{X,Y}$. Hence the transposition bijection identifies them, so $u$ is a unit for $\mathrm{comp}$.

Therefore the internal hom carries a natural composition morphism and objectwise unit morphisms satisfying associativity and the unit laws.