If $\alpha\in T_n$ and $\beta\in S_n$ show that $\alpha \mathcal{R} \alpha\beta$ ($T_n$ is the full transformation monoid, and $S_n$ is the symmetric group, both on $\{1,2,\ldots ,n\}$).
Does this mean show $\alpha T_n = (\alpha\beta )S_n$? How would you do it?
More generally, given a monoid $M$ with group of units (i.e. invertible elements) $G$, then for all $s \in M$ and $g\in G$, $s\ \mathcal{R}\ sg$, since $sgg^{-1} = s$.
Now $sgG = sG$ is in general different from $sM$. For instance, if $G = \{1\}$, and $g = 1$, then $sgG = \{s\}$, which is in general different from $sM$.
Would $\alpha T_n = (\alpha\beta )S_n$ then $\alpha T_n = \alpha S_n$ what generally speaking is not true.
Since $\alpha\beta \in \alpha T_n\beta \subset \alpha T_n$ and $\alpha \in \alpha\beta T_n\beta^{-1} \subset \alpha\beta T_n$ then $\alpha \mathcal{R} \alpha\beta$.
