Let $E$ be a finite semigroup. According to N. Bourbaki (Algèbre I p. 121 exerc. 14 c), if $M$ and $M'$ are minimal right ideals in $E$, then they are isomorphic. I spent some time browsing through Clifford and Preston's monography "the algebraic theory of semigroups", also reading a few articles by Rees, Clifford and Dubreil, and found absolutely no mention of this fact anywhere. Maybe I haven't looked hard enough. If there is any semigroup specialist, could you tell me what you think?
This is in CliffordPreston. First every minimal left (right) ideal is inside the minimal twosided ideal $I$ (which is unique), see Exercise 13 on page 84. Second, the ideal $I$ is a simple semigroup (obvious, but is also in CP). Third, by Sushkevich's theorem (Appendix A), all maximal subgroups of $I$ are isomorphic. By Theorem 1.27, every minimal left ideal $L$ in $I$ is a left group, i.e. the direct product of the maximal subgroup and a left zero semigroup (whose order does not depend of $L$). Hence all minimal left ideals are isomorphic.
I will use Green's relations, definitions are here if you don't know them : http://en.wikipedia.org/wiki/Green's_relations.
If you represent the smallest Dclass $C_\min$ by a box as in the wikipedia article, then the right (resp. left) minimal ideals are exactly the rows (resp. columns) of $C_\min$, i.e. its R and Lclasses. This can be seen by the fact that every right ideal must contain a Rclass of $C_\min$ (otherwise it would not be an ideal), and that it must also be included in a single Rclass of $C_\min$ to be minimal. Then it suffices to see that for all $e\in E$, if $R$ is a Rclass of $C_\min$ , then so is $eR$, and $x\mapsto ex$ is an isomorphism between $R$ and $eR$.

$\begingroup$ Thanks for your answer. Why should $x \mapsto ex$ be an isomorphism? This map is obviously oneone and onto but why do we have $exy=exey$ if $x,y \in R$? $\endgroup$ – Pnine Jul 3 '12 at 20:16