The right-module endofunctor category is strict monoidal

For objects $(F,c)$ and $(G,d)$ of ${\mathcal{C}}^{\mathrm{\prime }}$, define $(F,c)\otimes (G,d):=(FG,e)$ with $e_{X,Y}:=F(d_{X,Y})\circ c_{G(X),Y}$. The axioms from [L1] for $c$ and $d$ imply the same associativity and unit equations for $e$, so $(FG,e)$ is again an object of ${\mathcal{C}}^{\mathrm{\prime }}$. For morphisms $\theta :(F,c)\Rightarrow (F^{\mathrm{\prime }},c^{\mathrm{\prime }})$ and $\eta :(G,d)\Rightarrow (G^{\mathrm{\prime }},d^{\mathrm{\prime }})$, define $(\theta \otimes \eta {)}_X:={\theta }_{G^{\mathrm{\prime }}(X)}\circ F({\eta }_X)=F^{\mathrm{\prime }}({\eta }_X)\circ {\theta }_{G(X)}$. The equality is naturality, and the module-compatibility equation follows from the corresponding equations for $\theta$ and $\eta$.

Let $\mathbf{I}$ be the identity functor on $\mathcal{C}$ with structure map $i_{X,Y}=1_{X\otimes Y}$. Then $(\mathbf{I},i)$ is an object of ${\mathcal{C}}^{\mathrm{\prime }}$ and acts as a two-sided unit for the tensor just defined.

Composition of endofunctors is literally associative and unital, so $((F\otimes G)\otimes H)=FGH=F\otimes (G\otimes H)$ and $\mathbf{I}\otimes F=F=F\otimes \mathbf{I}$ as equalities of objects. The induced structure maps agree term by term from the definition in step 1.1, so the associator and unitors are identities.

Step 2.1 verifies the strictness clause of [L2], so ${\mathcal{C}}^{\mathrm{\prime }}$ is strict monoidal under composition.