Let $\mathcal{G},\mathcal{G}'$ be sheafs on a topological space $X$ such that $$\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{G},\mathcal{F})\cong\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{G}',\mathcal{F})$$ for all sheaves $\mathcal{F}\in\operatorname{Sh}(X)$. Then why does it follow that $\mathcal{G}\cong \mathcal{G}'$?
I understand this goes along the lines explicitly as: taking $\mathcal{F}=\mathcal{G}$, we get a morphism $\varphi:\mathcal{G}'\rightarrow\mathcal{G}$ corresponding the the identity on $\mathcal{G}$ from the above isomorphism; similarly, we have a morphism $\psi:\mathcal{G}\rightarrow\mathcal{G}'$ corresponding to the identity of $\mathcal{G}'$.
The claim is that from naturality of the morphisms, $\varphi$ and $\psi$ are inverses of each other, but I having trouble arriving to this conclusion. This is equivalent to saying that
I can't exactly come to terms why naturality gives the claim, though the diagram hints something along its lines. How does this follow? I will be really grateful for any help!
It helps to give a name to the natural isomorphism, say $$ \alpha_{\mathcal F} \colon \operatorname{Hom}_{\mathcal O_X}(\mathcal G,\mathcal F) \xrightarrow{\cong} \operatorname{Hom}_{\mathcal O_X} (\mathcal G', \mathcal F). $$ If $\sigma\colon \mathcal F\to \mathcal F'$ is a morphism, then naturality says $$ \alpha_{\mathcal F'}(\sigma\circ \rho) = \sigma\circ \alpha_{\mathcal F}(\rho), \qquad\text{in $\operatorname{Hom}_{\mathcal O_X}(\mathcal G', \mathcal F')$}, $$ for all $\rho\colon \mathcal G\to \mathcal F$.
So now, put $\varphi := \alpha_{\mathcal G}(\operatorname{id}_{\mathcal G}) \colon \mathcal G'\to \mathcal G$ and let $\psi \colon \mathcal G\to \mathcal G'$ be the unique morphism with $\alpha_{\mathcal G'}(\psi) = \operatorname{id}_{\mathcal G'}$. We then compute $$ \psi\circ \varphi= \psi \circ \alpha_{\mathcal G}(\operatorname{id}_{\mathcal G}) =\alpha_{\mathcal G'}(\psi\circ \operatorname{id}_{\mathcal G}) = \alpha_{\mathcal G'}(\psi) = \operatorname{id}_{\mathcal G'} $$ and $$ \alpha_{\mathcal G}(\varphi\circ \psi) = \varphi \circ \alpha_{\mathcal G'}(\psi) = \alpha_{\mathcal G}(\operatorname{id}_{\mathcal G}) \circ \operatorname{id}_{\mathcal G'} = \alpha_{\mathcal G}(\operatorname{id}_{\mathcal G}). $$ Since $\alpha_{\mathcal G}$ is injective, we deduce $\varphi\circ\psi = \operatorname{id}_{\mathcal G}$. Hence, $\varphi$ and $\psi$ are inverse to each other.