Yoneda lemma - Cayley’s theorem and Example with cyclic group $\mathbb{Z}_6$. I want to see in the specific the case $\mathbb{Z}_6$. From the prospective of category theory the group $\mathbb{Z}_6$ is a category with one object $X$ and the isomorphisms from $X$ to itself are the elements of $\mathbb{Z}_6$.
From the Yoneda lemma, since $|\mathbb{Z}_6|$ = 6, I'm expecting the set of natural transformations (between the $\hom$-sets) to have order $6$ as well.
Now, assuming what I’ve said so far to be correct, my issue is related to the use of the automorphism group in the context of Cayley’s theorem. Specifically I read here, bottom of the page:
\begin{align*}
  \mathrm{Nat}(\hom(-, \cdot), \hom(-, \cdot))
  &\cong \hom(\cdot, \cdot) \,,
  \qquad
  \text{(the lemma)}
  \\
  \mathrm{Aut}(G) &\cong G \,,
\end{align*}
which in the case $G=\mathbb{Z}_6$ would give $\mathrm{Aut}(\mathbb{Z}_6) \cong \mathbb{Z}_6$, which is not correct since $\mathrm{Aut}(\mathbb{Z}_6) = \mathbb{Z}_2$.
Is it because there are two different definitions of $\mathrm{Aut}(G)$? One from ordinary group theory and one from category theory? Category theory states that for a given object $X$ in $C$ then $\mathrm{Aut}(X)$ is the collection of isomorphisms of $X$. This definition and the Yoneda lemma would indicate there are six natural transformations and not only $2 = |\mathbb{Z}_2|$.
I’m ok with Cayley’s theorem that any group $G$ is isomorphic to a subset of $\mathrm{Sym}(G)$, but it’s the use of $\mathrm{Aut}(G)$ which confuses me. Most of the material on the introduction to category theory (S Roman, P Smith, F Lawvere and S Schanuel) explain the relation between the Yoneda Lemma and Cayley’s theory but none of the mention the $\mathrm{Aut}(G)$. I only found in ‘Category Theory in Context‘ - E Riehl (page 61) a brief mention of $\mathrm{Aut}(G)$ but there are no enough details (for me) and no examples.
Thank you for any info.
 A: $\operatorname{Aut}(G)$ should indeed be $G$, as the natural transformations are induced by elements of $G$ and not group automorphisms $G\rightarrow G$. To take it apart in detail:
Category theoretic part: Take a group $G$ and the corresponding groupoid (category, in which every morphism is invertible) with one object $\bullet$ and $\operatorname{End}(\bullet)=G$, often denoted $\mathbf{B}G$. A natural transformation $\eta\colon\operatorname{Hom}(-,\bullet)\Rightarrow\operatorname{Hom}(-,\bullet)$ of the (covariant) functor $\operatorname{Hom}(-,\bullet)\colon\mathbf{B}G^\mathrm{op}\rightarrow\mathbf{Grp}\rightarrow\mathbf{Set}$ to itself now in particular consists of a family of morphisms $\eta_X\colon\operatorname{Hom}(X,\bullet)\rightarrow\operatorname{Hom}(X,\bullet)$ for every object $X$ of $\mathbf{B}G^\mathrm{op}$, but there is only one such object, which is $X=\bullet$, and therefore $\eta$ consists of only one map $\eta_\bullet\colon G\rightarrow G$ (Be aware, that we don't know, if it is a group automorphism!). For all morphisms $g\colon X\rightarrow Y$ in $\mathbf{B}G^\mathrm{op}$, we also need, that $\eta_Y\circ\operatorname{Hom}(g,\bullet)=\operatorname{Hom}(g,\bullet)\circ\eta_X$, but again $X=Y=\bullet$ and furthermore $g\colon\bullet\rightarrow\bullet$ resembles an element of $G$, and therefore $\eta_\bullet\circ\operatorname{Hom}(g,\bullet)=\operatorname{Hom}(g,\bullet)\circ\eta_\bullet$ for every $g\in G$.
Group theoretic part: $\operatorname{Hom}(g,\bullet)$ is given by precomposition with $g$ and composition in $\mathbf{B}G^\mathrm{op}$ is given by the dual of the group operation $\circ$ of $G$. Therefore $\operatorname{Hom}(g,\bullet)$ is the composition with $g$ from the right (which is what is meant with "right $G$-set" in the explanation you have linked). The above equation therefore means, that $\eta_\bullet\colon G\rightarrow G$ is a map, for which $\eta_\bullet(h\circ g)=\eta_\bullet(h)\circ g$ for every $g,h\in G$. By putting $h=e$ as the neutral element $e\in G$, we get, that $\eta_\bullet(g)=\eta_\bullet(e)\circ g$, so $\eta_\bullet$ is the composition with $\eta_\bullet(e)\in G$ from the left. As one can check in reverse, every map $G\rightarrow G$ given by composition with an element $g\in G$ from the left (or from the right, switching from the dual of the group operation back to the group operation, which you see as a map $f_g\colon G\rightarrow G,x\mapsto xg$ in the explanation you have linked), fulfills the above condition and therefore defines a natural transformation.
Here we have the problem: "And these are precisely automorphisms of
$G$, that arise from multiplication by a fixed element!" But these maps $f_g$ are not even group homomorphisms, if the element $g\in G$ is not the neutral element. But they are indeed bijective with the inverse map $f_g^{-1}$ given by the map $f_{g^{-1}}$ corresponding to the inverse element.
Grand scheme of things: There is a canonical correspondence between natural transformations $\eta\colon\operatorname{Hom}(-,\bullet)\Rightarrow\operatorname{Hom}(-,\bullet)$, which are maps $G\rightarrow G$ given by composition with an element $g\in G$ from the left, and elements of $\operatorname{Hom}(\bullet,\bullet)=G$, just as the Yoneda lemma claims.
