Let $G$ be a group with neutral element $e$ and composition law $\ast$.

Let $G^{\rm \,op}$, called the opposite to $(G, \ast)$, be the group whose set of elements equals the set $G$, but whose group composition law $\bar{\ast}$ is $h\bar{\ast}g:= g\ast h$.

If $G$ is abelian, then $(G,\ast)=(G^{\rm \,op},\bar{\ast})$, so we assume in this question that $G$ is **not necessarily** abelian.

Let ${\rm Aut}(G)$ be the group of automorphisms of $G$, with composition law $\circ$ given by composition of maps.

An action of a group $(G, \ast)$ on a set $X$ is a map $g\mapsto \alpha_g$ from $G$ to the set of bijective maps from $X$ to $X$, such that $\alpha_e(x)=x$, for all $x\in X$, and $\alpha_{h\ast g}=\alpha_h\circ\alpha_g$, all $h,g\in G$, where $\circ$ is composition of maps.

For $g\in G$, let $L_g:G\rightarrow G$ be the map $L_g(h)=g\ast h$, let $R_g:G\rightarrow G$ be the map $R_g(h)=h\ast g$, and let ${\rm Ad}_g$ be the map ${\rm Ad}_g(h)=g\ast h\ast g^{-1}$, for all $h\in G$.

We call the $L_g$ left multiplications, the $R_g$ right multiplications, and the ${\rm Ad}_g$ adjoint maps, a.k.a. conjugations.

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