Yoneda lemma as a generalisation of Cayley's theorem I have seen answers in questions asking the same question. They have first described what is Yoneda lemma and then deduced Cayley's theorem from that. I am not asking for that.
I am planing to explain Yoneda lemma for  a group of students who know some group theory some basic definitions in category theory.
I don't want to prove Yoneda lemma and say that as a special case we can get Cayley's theorem. I want to recall Cayley's theorem and then say that it can be generalised to some extent and then state and prove Yoneda lemma.
Any suggestions are welcome.
 A: $\newcommand{\Sym}{\operatorname{Sym}}$
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$\newcommand{\Ar}{\operatorname{Ar}}$
$\newcommand{\Hom}{\operatorname{Hom}}$
$\newcommand{\End}{\operatorname{End}}$
$\newcommand{\Nat}{\operatorname{Nat}}$
$\newcommand{\cod}{\operatorname{cod}}$
$\newcommand{\dom}{\operatorname{dom}}$
$\newcommand{\id}{\operatorname{id}}$
$\newcommand{\com}{\operatorname{com}}$
$\newcommand{\Set}{\mathrm{Set}}$
First of all, I personally define categories to be structures $C = (\Ob C, \Ar C, \dom, \cod, \id, ∘)$, where $\Ob C$ is the class of objects, $\Ar C$ is the class of arrows and

*

*$\dom \colon \Ar C → \Ob C$ is the domain.

*$\cod \colon \Ar C → \Ob C$ is the codomain.

*$\id \colon \Ob C → \Ar C$ is the identity selection.

*$∘ \colon \Ar C × \Ar C \dashrightarrow \Ar C$ the arrow composition.

Adapt to your definition. This what I would do then:
Let $G$ be a group. Cayley says:

There is an injective group homomorphism $G → \Sym G$, where $\Sym G$ is the symmetric group on $G$, that is the group of bijective maps $G → G$.
Proof. The injection is given by $G → \Sym G,~g ↦ (g·~)$ where
$(g·~)$ is the left multiplication by $g$. It’s a homomorphism since
$∀g,h ∈ G \colon (gh·~) = (g·~)∘(h·~)$. It clearly is injective
because $(g·~) = \id_G ⇒ g·1 = 1$, so its kernel is trivial.

The proof also shows that the image of each $g ∈ G$, as a left multiplication, respects the right multiplication by $h ∈ G$ since $∀x ∈ G\colon g(xh) = (gx)h$. So we can specify Cayley:

There is an injective group homomorphism $G → \Nat G$, where $\Nat G$ is the subgroup of $\Sym G$ consisting of all bijective maps $G → G$ respecting the right action of $G$, i.e. $$\Nat G = \{σ ∈ \Sym G;~σ(xh) = σ(x)h\quad∀x,h ∈ G\}.$$

In fact, every such map $σ$ in $\Nat G$ must be the left multiplication by $σ(1)$, so this gives a bijection $G → \Nat G$. But let’s forget that for a second. We will soon see why we have named this subgroup $\Nat G$.
Now, let’s interpret $G$ as a category in the following way:

Let $C = (\Ob C, \Ar C, \dom, \cod, \id, ∘)$ be the category with $\Ob C = \{\star\}$, $\Ar C = G$ with $\cod = \dom = \star$ (constantly) and $\id \star = 1$ and $∘$ is the group multiplication $G × G → G$.

Now consider the contravariant hom-functor $h_\star = \Hom_C (–,\star) \colon C → \Set$. The only object it can accept is $\star$ and $\Hom_C (\star,\star) = \Ar C = G$. Any natural transformation $h_\star → h_\star$ is therefore given by a sole arrow $\Hom_C (\star,\star) → \Hom_C(\star,\star)$, i.e. by a map $G → G$ (since in general it’s given by a family of arrows – one for each object in $C$ – but here there’s only one object). This map $σ\colon G → G$ also respects the right action of $G$ – this comes from the naturality: For $x ∈ G$ and $h ∈ H$ we have
$$σ(xh) = (σ∘h_\star(h))(x) = (h_\star(h)∘σ)(x) = σ(x)h,$$
and vice versa: By using the very same equation we see that very such map $σ$ respecting the right action of $G$ automatically yields a natural transformation $h_\star → h_\star$.
Thus, $\Nat (h_\star, h_\star) \cong \Nat G$ as monoids. And $G = \Ar C = \Hom_C (\star,\star)$. And we can restate Cayley as

There is an injective monoid homomorphism $\Hom_C(\star,\star) → \Nat (h_\star, h_\star)$.

Because $\star$ is the only object of $C$ and $\Nat (h_\star,h_\star) = \Hom_{\Set^{C^\mathrm{op}}} (h_\star,h_\star) ⊆ \Ar \Set^{C^\mathrm{op}}$, one can enlarge such a monoid homomorphism to a functor by sending the object $\star$ to the object $h_\star$, so Cayley can be further restated as

There is a faithful functor $C → \Set^{C^\mathrm{op}}$.

It turns out, the last statement even holds true when $C$ is any category, not just a category that is a group in disguise. And even more is true: This functor is in fact also full (which is just a generalization of the fact that the Cayley embedding restricts to a bijection $G → \Nat G$, which we have already seen above).
Then I would state some version of Yoneda …
