See here.
In the category of $G$-sets, the morphisms $f:G/H\to X$ are in one-to-one correspondence with the elements of $X^H$; the correspondence sends $f$ to $f(H)$ (where the subgroup $H$, being a coset of itself, is an element of $G/H$). After checking that this correspondence is natural...
Sorry, but how do we show naturality here? It's not clear to me...
Let $\phi_H: \textbf{Hom}_{G-set}(G/H,X) \to X^H$ be $\phi_H(f)=f(H)$. Let $\alpha :G/H' \to G/H$ be an equivariant map. To prove naturality we want to show that $$\textbf{Hom}_{G-set}(G/H,X) \overset{\phi_H}{\longrightarrow} X^H \overset{X^{\alpha}}{\longrightarrow} X^{H'}$$ and $$ \textbf{Hom}_{G-set}(G/H,X) \overset{\alpha^*}{\longrightarrow} \textbf{Hom}_{G-set}(G/H',X) \overset{\phi_{H'}}{\longrightarrow} X^{H'} $$ are the same map. Note that $X^{\alpha}(x) = \gamma x$ where $\alpha(H')=\gamma H$. Following the first diagram we get $X^{\alpha}(\phi_H(f)) = \gamma f(H) $ and following the second diagram we get $\phi_{H'}(\alpha^*(f)) = \phi_{H'}(f\alpha) = f(\alpha(H')) = f(\gamma H) = \gamma f(H)$. Thus the transformation is natural. It's also a bijection as noted at the link making it a natural isomorphism between $X^{(-)}$ and the representable functor $\textbf{Hom}_{G-set}(-,X)$.
This problem is verbatim from Peter May's book A Concise Course in Algebraic Topology.