For $G$ a group, the left multiplication $a\mapsto(g\mapsto ag)$ and the right multiplication by the inverse $a\mapsto(g\mapsto ga^{-1})$ provide two embeddings of $G$ into $\operatorname{Sym}(G)$, of image sets $\Lambda$ and $R$, respectively. Some facts about $\Lambda$ and $R$ are as follows.
$$\Lambda R\le\operatorname{Sym}(G)\tag1$$
In fact, $(\lambda_a\rho_b)(g)=$ $\lambda_a(\rho_b(g))=$ $\lambda_a(gb^{-1})=a(gb^{-1})=$ $(ag)b^{-1}=$ $\rho_b(ag)=$ $\rho_b(\lambda_a(g))=$ $(\rho_b\lambda_a)(g)$, whence $\lambda_a\rho_b=\rho_b\lambda_a$, and hence $\Lambda R=R\Lambda$.
$$\Lambda\cap R\cong Z(G)\tag2$$
In fact, $\lambda_a\in R\iff$ $\exists b\in G:\lambda_a=\rho_b\iff$ $\exists b\in G:ag=gb^{-1}$, for all $g\in G\Longrightarrow$ $b=a^{-1}\Longrightarrow$ $\lambda_a=\rho_{a^{-1}}\iff$ $ag=ga$ for all $g\in G\iff$ $a\in Z(G)$.
$$\operatorname{Inn}(G)\le\Lambda R\tag3$$
In fact, $\operatorname{Inn}(G)=\{\lambda_a\rho_{a}, a\in G\}$.
$$\Lambda\cap\operatorname{Aut}(G)=R\cap\operatorname{Aut}(G)=1\tag4$$
In fact, $\lambda_a\in\operatorname{Aut}(G)\iff$ $\forall g,h\in G:agh=agah\iff$ $1=a\iff$ $\lambda_a=1$; likewise, $\rho_b\in\operatorname{Aut}(G)\iff$ $\forall g,h\in G:ghb^{-1}=gb^{-1}hb^{-1}\iff$ $1=b^{-1}\iff$ $b=1\iff$ $\rho_b=1$.
Next:
$$\operatorname{Inn}(G)=\Lambda R\cap\operatorname{Aut}(G)\tag5$$
In fact, $\lambda_a\rho_b\in\operatorname{Aut}(G)\iff$ $\forall g,h\in G:(\lambda_a\rho_b)(gh)=(\lambda_a\rho_b)(g)(\lambda_a\rho_b)(h)\iff$ $aghb^{-1}=agb^{-1}ahb^{-1}\iff$ $1=b^{-1}a\iff$ $b=a\iff$ $\lambda_a\rho_b=\lambda_a\rho_a\in\operatorname{Inn}(G)$.
Finally, denoted with $\operatorname{Eqv}(G)$ the group of the equivariant maps of $G$, we get the following facts.
$$\operatorname{Eqv}(G)\cap\Lambda R=\Lambda\cap R\tag6$$
In fact, $\lambda_a\rho_b\in\operatorname{Eqv}(G)\iff$ $(\lambda_a\rho_b)(gh)=g((\lambda_a\rho_b)(h))\iff$ $aghb^{-1}=gahb^{-1}\iff$ $ag=ga$ for every $g\in G\iff$ $a\in Z(G)$.
$$\operatorname{Eqv}(G)\cap\operatorname{Aut}(G)=1\tag7$$
In fact, $f\in\operatorname{Aut}(G)\cap\operatorname{Eqv}(G)\Longrightarrow$ $af(g)=f(a)f(g)\iff$ $f(a)=a, \forall a\in G\iff$ $f=1$.
A possible (admittedly amongst the infinitely many) "consistent" visual representation of $(2)\div(7)$ would go as follows:
My question is: does this naive representation convey any real mathematical fact about the involved sets? In particular, are there (topological?) groups $G$ where the "separation" of $1$ from the rest of $\operatorname{Aut}(G)$, and of $\Lambda\cap R$ from the rest of $\operatorname{Eqv}(G)$, can be made precise in some mathematical sense? (Btw, note that both "separations" persist in the case of abelian $G$.)