Let $S$ be a finite semigroup, two elements $s,t \in S$ are called $\mathcal R$-equivalent if $s = t$ or there exists $x,y \in S$ such that $s = tx, t = sy$. Equivalently they generate the same right ideal, i.e. $$ sS \cup \{s\} = tS \cup \{t\}. $$ For more information see Green's relations.
Now if two idempotents are $\mathcal R$-equivalent, are they equal, i.e. if $e \mathcal R f$ does this imply $e = f$ in finite semigroups?
In general two idempotents $e,f$ are $\mathcal R$-equivalent if and only if $ef = f$ and $fe = e$, as $ex = f$ implies $f = ex = eex = ef$ and similary $fe = e$. If we denote by $E(S)$ the idempotents of $S$, then $e,f \in E(S)$ are $\mathcal R$-equivalent in $S$ iff they are $\mathcal R$-equivalent in $E(S)$.
The answer is no. Consider the two-element semigroup $\{e, f\}$ defined by $ee = e$, $ff = f$, $ef = f$ and $fe = e$.