= Solution
A <diagram in a category> is a functor $D:\mathcal J\to\mathcal C$. A <cone over a diagram> with vertex $A$ is a family
$$
\gamma_j:A\to D(j)
$$
such that $D(u)\gamma_i=\gamma_j$ for every $u:i\to j$. A <categorical limit> is a terminal cone: for every cone $(A,\gamma)$ there is a unique map $A\to\lim D$ commuting with all legs.
Suppose $\mathcal C$ has small <product in a category>[products] and <equalizer>[equalizers]. For a small diagram $D$, form
$$
P=\prod_{j\in\mathcal J}D(j),
\qquad
Q=\prod_{u:i\to j}D(j).
$$
There are two maps $r,s:P\rightrightarrows Q$. In the coordinate indexed by $u:i\to j$, let
$$
r_u=D(u)\pi_i,
\qquad
s_u=\pi_j.
$$
The <equalizer> $E\to P$ imposes exactly the cone equations. Maps $A\to E$ are therefore naturally the same as cones from $A$ to $D$, so $E\cong\lim D$. This is the <construction of small limits from products and equalizers>.
Let $F:\mathcal I\to\mathcal J$ be <initial functor>[initial], so every $(F\downarrow j)$ is nonempty and <connected category>[connected]. Restriction sends a cone $(A,\gamma_j)$ over $D$ to $(A,\gamma_{Fi})$ over $DF$. Conversely, given a cone $(A,\delta_i)$ over $DF$, choose an object
$$
(i,u:Fi\to j)\in(F\downarrow j)
$$
and define
$$
\gamma_j=D(u)\delta_i.
$$
A morphism in the comma category shows that this expression is unchanged along one edge, and connectedness makes it independent of the chosen object. The cone equations follow by choosing $(i,vu)$ for an arrow $v:j\to j'$. This construction is inverse to restriction and acts identically on vertex maps, proving the <cone restriction along an initial functor> isomorphism.
Terminal objects in the two cone categories therefore correspond. Whenever the $\mathcal I$-shaped limit exists,
$$
\lim_{\mathcal J}D\cong\lim_{\mathcal I}DF
$$
naturally in $D$. Equivalently, the triangle formed by precomposition
$$
F^*:[\mathcal J,\mathcal C]\to[\mathcal I,\mathcal C]
$$
and the two limit functors commutes up to natural isomorphism.
For the converse, suppose this commutation holds for $\mathcal C=\mathbf{Set}^{\mathrm{op}}$. Passing to <opposite category>[opposite categories] says that restriction along $F^{\mathrm{op}}$ preserves all set-valued colimits. Fix $j\in\mathcal J$ and take the representable functor
$$
H=\mathcal J(-,j):\mathcal J^{\mathrm{op}}\to\mathbf{Set}.
$$
Its colimit is a singleton: the category of its elements has the initial object $(j,1_j)$. The restricted colimit is
$$
\operatorname*{colim}_{i\in\mathcal I^{\mathrm{op}}}\mathcal J(Fi,j),
$$
whose elements are precisely the connected components of $(F\downarrow j)$. By the assumed comparison this set is also a singleton. Thus $(F\downarrow j)$ is nonempty and connected for every $j$, so $F$ is initial.
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