$\newcommand{\A}{\mathscr{A}}\newcommand{\Gc}{\mathscr{G}}\newcommand{\G}{\mathcal{G}}\newcommand{\s}{\mathsf{Set}}\newcommand{\op}{^{\mathsf{op}}}\newcommand{\sym}{\mathsf{Sym}}$I am having difficulty applying the Yoneda lemma to Cayley's theorem. I have just learnt and understood the proof of the Yoneda lemma, in its presheaf/contravariant form. A preamble on the notation that I use and the context for what I know:
Let $\A$ be any locally small category. We have that: $$[\A\op,\s](H_A,X)\cong X(A)$$Naturally in $(A,X)\in\A\op\times[\A\op,\s]$, where $H_A:\A\op\to\s$ is the functor defined on objects $B\in\A$ by $B\mapsto\A(B,A)$ and on arrows $f\in\A(B',B)$ by $f\mapsto -\circ f:\A(B,A)\to\A(B',A)$.
In the contravariant version, the forward isomorphism, $(\hat{})_A:[\A\op,\s](H_A,X)\to X(A)$ is the mapping $\alpha\mapsto\alpha_A(1_A)$.
In the covariant version, we consider, for $A\in\A$, the functor $H^A:\A\to\s$ which maps objects $B\mapsto\A(A,B)$ and arrows $f\in\A(B,B')$ by $f\mapsto f\circ-:\A(A,B)\to\A(A,B')$.
I like this form of the lemma since it talks about presheaves, which have their own, more compact notation of $\mathsf{Psh}_{\A}$. Anyway, the lemma of course dualises. I am concerned with the proof of Cayley's theorem that I saw on the relevant Wikipedia page. Let's follow it through until the problem is reached:
Let $\G$ be a group. $\G$ has a categorical implementation as a one-object category, $\Gc$, with object $\{\ast\}$, where all arrows $\{\ast\}\to\{\ast\}$ are isomorphisms and the group action is arrow composition. Any functor $F:\Gc\to\s$ results in a set $S:=F(\ast)$ and all elements $g\in\G$, identified with the arrows in $\Gc$, result in functions $F(g):S\to S$. If we denote $g\cdot s:=F(g)(s)$, then the left action of $\G$ on $S$ makes $S$ a left $\G$-set. Natural transformations $\alpha:F\implies F$ result in equivariant maps: $\alpha(g\cdot x)=\alpha(F(g)(x))=F(g)\alpha(x)=g\cdot\alpha(x)$ by naturality.
Let $X:=H^{\ast}$ in the dual version of the Yoneda lemma. Then $[\Gc,\s](H^\ast,H^\ast)\cong H^\ast(\ast)$, so all equivariant maps on the left $\G$-set $G:=\Gc(\ast,\ast)$ correspond naturally to the elements $g\in G$. It remains to show that the equivariant maps form a subgroup of $\sym(\G)$ under composition and that the isomorphism is a group isomorphism.
This is where I am unsure.
Problem 1: I don't know how to sure the equivariant maps are invertible; the obvious try is to consider $\alpha(1)=\alpha(g\cdot g^{-1})=g\cdot\alpha(g^{-1})\implies g=\alpha(1)\cdot(\alpha(g^{-1}))^{-1}$. I am not sure how to make an equivariant map $\beta$ which uses the above principle to invert $\alpha$.
Problem 2: It is clear that $(\hat{})$ preserves the identity. To show it is a group isomorphism I need only show that $\hat{(\beta\circ \alpha)}=\hat{\beta}\cdot\hat{\alpha}$. However, $\hat{(\beta\circ\alpha)}=\beta(\alpha(1))=\beta(\alpha(1)\cdot1)=\alpha(1)\cdot\beta(1)=\hat{\alpha}\cdot\hat{\beta}$ which is the wrong way round. In my notes I rather sloppily said "Oh it's dual" so I reversed it to obtain $\hat{\beta}\cdot\hat{\alpha}$ but that isn't good enough; in particular, the Wikipedia page cites the same isomorphism $(\hat{})$ for the covariant version of the lemma as well, so I don't know what's going wrong.
Many thanks for any clarifications to problems 1 or 2 (preferably both).
It is my conclusion that Wikipedia is just obstinate on the use of the covariant version, introducing further confusions to students like myself. Every group (-category) is isomorphic to its dual (-category) so the issue with $\hat{(\beta\circ\alpha)}=\hat{\alpha}\cdot\hat{\beta}$ being “the wrong way round” is easily addressed by passing to the dual. Alternatively, and much more clearly, we can use the standard contravariant “presheaf” version, and now the equivariant functions are right-equivariant, introducing the correct order without needing to pass to the dual, although passing to the dual is exactly the process of showing equivalence between the covariant and contravariant version.
We see that the equivariant maps are invertible by, e.g. for left-maps, $\alpha(g1)=g\alpha(1)$ so $\beta:g\mapsto g\cdot(\alpha(1))^{-1}$ is a suitable inverse.