Group as a category with one object

Let $G$ be a group. We are going to construct a category with one element whose morphisms are elements of this group $G$. Definition : A category $\mathcal{C}$ consists of

a collection of objects, $\text{Ob}(\mathcal{C})$,
for each $A,B\in \text{Ob}(\mathcal{C})$ a collection of maps from $A$ to $B$ denoted by $\mathcal{C}(A,B)$.
for each $A,B,C \in \text{Ob}(\mathcal{C})$ a function $\mathcal{C}(B,C)\times \mathcal{C}(A,B)\rightarrow \mathcal{C}(A,C)$ with $(g,f)\mapsto g\circ f$ called the composition
for each $A\in \text{Ob}(\mathcal{C})$ an element $1_A\in \mathcal{C}(A,A)$ called the identity on $A$

satisfying the following conditions

Associativity : for each $f\in \mathcal{C}(A,B), g\in \mathcal{C}(B,C), h\in \mathcal{C}(C,D)$ we have $(h\circ g)\circ f=h\circ (g\circ f)$.
Identity law : for each $f\in \mathcal{C}(A,B)$ we have $f\circ 1_A=f=1_B\circ f$.

We are not constructing a categroy $\mathcal{C}$ with $\text{Ob}(\mathcal{C})=G$, we are constructing a category $\mathcal{C}$ with $\text{Ob}(\mathcal{C})=\{A\}$(one point set) and $\mathcal{C}(A,A)=G$. We have defined objects and morphisms collections (here there is only one object so there is only one morphisms collection). We have to define composition $\mathcal{C}(A,A)\times \mathcal{C}(A,A)\rightarrow \mathcal{C}(A,A)$ i.e., we have to give a map $G\times G\rightarrow G$. There are two natural ways to give this map $(g,h)\mapsto g.h$ or $(g,h)\mapsto h.g$. We consider the map $(g,h)\mapsto g\circ h:=g.h$ to give composition. Obvious choice of an identity element in $\mathcal{C}$ is $llatex 1_A=e$ identity element of the group. Associativity of group implies associativity of composition

$(g\circ h)\circ f=(g.h)\circ f=(g.h).f=g.(h.f)=g\circ (h\circ f).$

We have

$1_A\circ g=e.g=g$ and $g\circ 1_A=g.e=g$.

So, we have a category whose sets of objects contains exactly one element and whose set of morphisms is exactly one set equal to given group $G$. Given a category $\mathcal{C}$ and an element $A\in\text{Ob}(\mathcal{C})$ we have a functor $h_A:\mathcal{C}^{\text{op}}\rightarrow \text{Sets}$ defined by $B\mapsto \mathcal{C}(B,A)$ and given an element $f\in \mathcal{C}(B,C)$ we have $h_A(f)\in \text{Mor}((h_A(C),h_A(B))$ given by $h_A(f)(g)=g\circ f$. We will see explicitly what is $h_A$ in case of group as a category with exactly one object as defined above. We have $h_A(A)=\mathcal{C}(A,A)=G$. Let $h\in \mathcal{C}(A,A)=G$. Then, $h_A(h):G\rightarrow G$ given by $g\mapsto g\circ h=g.h$. Let us see explicitly what are the natural transformations $\eta:h_A\rightarrow h_A$. This should come up with $\eta(A):h_A(A)\rightarrow h_A(A)$ i.e., with a map $\eta(A)=\sigma:G\rightarrow G$ satisfying the naturality condition i.e., for every $h\in G=\mathcal{C}(A,A)$ the following diagram commutes ql_5fecb8d9fff6db2810f2e1d2eec2e7a6_l3

i.e., we want $\sigma:G\rightarrow G$ to be such that $(h_A(h)\circ \sigma)(g)=(\sigma\circ h_A(h))(g)$ for all $g\in h_A(A)=G$ i.e., $h_A(h)(\sigma(g))=\sigma(h_A(h)(g))$ ie., $\sigma(g)h=\sigma(gh)$ for all $g\in G$. So, the set of all natural transformations from $h_A$ to $h_A$ is given by

$\text{Nat}(h_A,h_A)=\{\sigma:G\rightarrow G : \sigma(gh)=\sigma(g)h \forall g,h\in G\}$.

See that every element in $\text{Nat}(h_A,h_A)$ is an isomorphism, as a set. It is not even necessary that $\sigma\in \text{Nat}(h_A,h_A)$ is a group homomorphism. See that $\sigma(h)=\sigma(1.h)=\sigma(1).h ~~\forall h\in G$. Given $\sigma\in \text{Nat}(h_A,h_A)$ define $\tau:G\rightarrow G$ as $\tau(h)=(\sigma(1))^{-1}h$. Then,

$\sigma(\tau(h))=\sigma(\sigma(1)^{-1}h)=\sigma(1).\sigma(1)^{-1}.h=h$.
$\tau(\sigma(h))=\tau(\sigma(1)h)=\sigma(1)^{-1}.\sigma(1).h=h$.

So, $\tau$ is inverse of $\sigma$. Thus, every element in $\text{Nat}(h_A,h_A)$ is an isomorphism of sets. Thus, we have

$\text{Nat}(h_A,h_A)=\{\sigma\in \text{Sym}(G) : \sigma(gh)=\sigma(g)h \forall g,h\in G\}\leq \text{Sym}(G)$.

Yoneda lemma states that $\mathcal{C}(A,A)\cong \text{Nat}(h_A,h_A)$ i.e., $G\cong H\leq \text{Sym}(G)$ which is exactly the statement of Cayley's theorem. So, yoneda lemma does say that any group can be identified with a subgroup of its symmetric group.