 Solution

Choose for each $j\in \mathcal{J}$ a categorical limit $L_j$ of $F(-,j)$, with projections ${\lambda }_{i,j}:L_j\to F(i,j)$. For $v:j\to k$, the morphisms $F(1_i,v){\lambda }_{i,j}$ form a categorical cone over $F(-,k)$, since the two actions of the product functor $F$ commute. Define $L(v):L_j\to L_k$ by the equationsExistence and uniqueness follow from the universal property. The identity morphism of $L_j$ satisfies the equations for $v=1_j$, and $L(w)L(v)$ satisfies those for $wv$, so uniqueness gives $L(1_j)=1_{L_j}$ and $L(wv)=L(w)L(v)$. Hence $L:\mathcal{J}\to \mathcal{D}$ is a functor.

If $L_j^{\prime }$ with projections ${\lambda }_{i,j}^{\prime }$ are other choices, there is a unique isomorphism $u_j:L_j\to L_j^{\prime }$ carrying every projection to the corresponding projection. The same equations show $u_{k}L(v)=L^{\prime }(v)u_j$, because their composites with all ${\lambda }_{i,k}^{\prime }$ agree. Thus the choices give a unique natural isomorphism compatible with the chosen categorical limit categorical cones. Without the compatibility condition, uniqueness of an arbitrary natural isomorphism is not asserted.