= 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 equations
$$
\lambda_{i,k}L(v)=F(1_i,v)\lambda_{i,j}\qquad(i\in\mathcal I).
$$
Existence 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 $L'_j$ with projections $\lambda'_{i,j}$ are other choices, there is a unique <isomorphism> $u_j:L_j\to L'_j$ carrying every projection to the corresponding projection. The same equations show $u_kL(v)=L'(v)u_j$, because their composites with all $\lambda'_{i,k}$ agree. Thus the choices give \b[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.
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