Representable functors and their representing objects

Reading the chapter concerning the Representable functors in the Emily Riehls' book "Category in context" I found myself a bit stuck when trying to figure out a representing object for a given functor. Excepting the naturality condition, What kind of necessarily conditions or maybe heuristics can we apply while searching for such representing object. Even the simplest example where the representing object (singleton set) for the identity functor on $\mathcal{Set}$ is not totally trivial (at least for me).

Thanks for any proposal

I think In got the trick, at least for the identity functor.... Now the object representing the identity functor is trivially the singleton. Why?

Let $C$ be the representing object, a set. The components $\alpha$ of the natural isomorphism must be isomorphisms, bijections in this case. But clearly if $|C| > 1$ certainly we cannot have a bijection $\alpha: A \rightarrow Hom(C,A), \forall A$ (take a finite set $A$ for example). Then $|C|=1$ and $C$ is the singleton set.

Now I have to understand the representation of the forgetful functor $U: \mathcal{Grp} \rightarrow \mathcal{Set}$...

This one also now I get the trick. For a given group $A$, let say finite for the sake of simplicity, in order to have a bijection $\alpha : A \rightarrow HOM_{\mathcal{Grp}}(C,A)$ we must have at least the same number of group morphismes in $HOM_{\mathcal{Grp}}(C,A)$ as elements in $A$. We can be tempted to do mimic what we done for the identity functor and take the trivial group $\{1_C\}$ as representing object. But that doesn't work because in order to have a group morphism we must have $1_C \mapsto 1_A$. QWe must find a group $C$ such that $\exists \ x \in C$ and an arrow $x \mapsto a \in A$ which extends to a group morphisms. The right candidate is $C = \left(\mathbb Z,+\right)$ with the map $1 \mapsto a$. This arrow extends naturally to a groups morphism and we have as much morphisms as elements in $A$. The candidate for representing the forgetful functor $U : \mathcal{Grp} \rightarrow \mathcal{Set}$ is the additive group $\left(\mathbb Z,+\right)$.

You want to find a group $G$ such that the two functors $\operatorname{Hom}(G, \cdot)$ and $U$ are naturally isomorphic.

In intuitive terms, this means that, for any group $H$, there is a "natural" way to identify the elements of $H$ with group homomorphisms from $G$ to $H$.

Which group $G$ is a good candidate for that? Answer: $G = \Bbb Z$.

Why? Because any group homomorphism $\Bbb Z \rightarrow H$ is uniquely determined by the image of $1 \in \Bbb Z$, which can be chosen to be any element of $H$.

I hope this is enough for you to write down a proper answer. Note that a general functor may not be representable, and it may not be easy to prove that certain functors are representable (e.g. Picard functor).