Let $\beta : I^{op} \to \text{Set}$ be a functor and $X \in \text{Set}$.
I want to show that there is a natural isomorphism
$$ \text{Hom}_{\text{Set}}(X, \lim_{\longleftarrow} \beta) \xrightarrow{\sim} \lim_{\longleftarrow} \text{Hom}_{\text{Set}}(X, \beta) $$
Firstly, what do they mean by natural here, with respect to what functoriality? It is given that the functor $\beta'$ in the limit on the right is $\beta'(i) = \text{Hom}_{\text{Set}}(X, \beta(i))$.
Secondly, how do we establish the mapping? I know that if $a : X \to \lim\limits_{\longleftarrow} \beta$ in the category $\text{Set}$, then for $x \in X$, $a(x)$ is a natural map in $\text{Hom}_{I^{\wedge}}(\text{pt}_{I^{\wedge}}, \beta)$ such that for all $f : j \to i$ in $I$ we have $\beta(f)\circ a(x)_i = a(x)_j$.
The right side is the same thing except $\text{Hom}_{Set}(X, \beta(f)) \circ a(x)_i =a(x)_j$ .
So they appear to be the same sort of. Does $\text{Hom}_{\text{Set}}(X, g) = g$? If not, how do we show isomorphism?
The term 'natural isomorphism' really just means natural isomorphism: the two expressions on either side of the $\overset{\sim}{\longrightarrow}$ symbol are both functors $I^{\mathrm{op}} \to \mathbf{Set}$. With that said, the isomorphism is also natural in $X$, but I imagine they're not asking you to prove that.
Now $\mathrm{Hom}_{\mathbf{Set}}(X, {-})$ is a functor $\mathbf{Set} \to \mathbf{Set}$. The action of this functor on functions is given by postcomposition. Explicitly, for a function $g : Y \to Z$, the function $$\mathrm{Hom}_{\mathbf{Set}}(X,g) : \mathrm{Hom}_{\mathbf{Set}}(X, Y) \to \mathrm{Hom}_{\mathbf{Set}}(X,Z)$$ is defined by $\mathrm{Hom}_{\mathbf{Set}}(X,g)(h) = g \circ h$ for all $h : X \to Y$.
Use this to (even more carefully) write down the action of both functors on objects of $I$. The fact that the left and right sides are (almost) 'the same thing' is what gives you the natural isomorphism!