Preservation of Limit by Hom: Naturality Question. $\DeclareMathOperator{\Hom}{Hom}$Let $D_{i}$ be a diagram in a category $C$,with d the limit. We have 
(1) $\lim \Hom_{C}(X,D_{i}) \cong \Hom_{C}(X,d)$.
Whenever I see this result used, it is always stated that the isomorphism is natural in $X$. I have never seen a proof so I want to do it from scratch. Here is my strategy:
I take $\lim : C^{I} \rightarrow C$ to be the functor defined on objects in the obvious way, and on arrows $D \rightarrow D'$ to be the unique morphism from d to d', that results from constructing the cone from d to $D_{i}'$. 
So LHS of (1) is $\lim \circ \Hom_{C}(-, D_{i}) : C^{op} \rightarrow Set$
Let $\lim \Hom_{C}(X, D_{i}) = S(X)$. Then letting $f : X' \rightarrow X$, I get a square whose top arrow is the iso $\phi : S(X) \rightarrow \Hom_{C}(X,d))$ and whose bottom arrow is the iso $\phi' : S(X') \rightarrow \Hom_{C}(X',d))$. The left arrow is the unique map from $S(X) \rightarrow S(X')$ and the right arrow is just the map between the two corresponding $\Hom_{C}$ functors.
But now I can't get the squares to commute. 

Suppose that $\{D_i, \psi_i^j\}$ is your system, and for $i \leq j$ you have a map $\psi_i^j : D_j \to D_i$, and let $\alpha_i : lim D_i \to D_i$ be the maps of the limit (the ones with the universal property). Then you have for every $X$ an isomorphism:
$$ \begin{array}{cccc} \theta_X : & Hom(X, d)  & \to & lim Hom(X,D_i) \\
                                &    \varphi & \mapsto & (\alpha_i \circ \varphi)_{i}       \end{array} $$
Let $f:X \to Y$, then you have induced maps $f^* :Hom(Y,d) \to Hom(X,d)$ and
$$ \begin{array}{cccc}  \overline{f^*} : & lim Hom(Y,D_i) & \to & lim Hom(X,D_i)\\
                          & (\varphi_i)_i & \mapsto & (\varphi_i \circ f)_i     \end{array} $$
Now for every $ \varphi \in Hom(Y,d) $ we have:
$$ \begin{array}{lcl} \overline{f^*} \circ \theta_Y (\varphi) & = & \overline{f^*} \big( (\alpha_i \circ \varphi)_i \big) \\ & = & (\alpha_i \circ \varphi \circ f)_i      \end{array} $$
And in the other hand:
$$ \begin{array}{lcl} \theta_X \circ f^* (\varphi) & = & \theta_X(\varphi \circ f) \\
                       & = &  (\alpha_i \circ \varphi \circ f)_i   \end{array} $$
And that's the naturality.
