Can essentially primitive idempotents be defined both as $e_\lambda=s_\lambda a_\lambda$ and $e_\lambda=a_\lambda s_\lambda$?

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I've been studying representation theory of symmetric group on Tung's Group Theory in Physics. I understood that different Young Diagrams corresponds to inequivalent irreducible representations of symmetric group (Theorem 5.6). In fact, they exhaust all irreducible representations (Theorem 5.7) and distinct standard Young Tableaux are linearly independent and the direct sum of left ideals generated by them span the whole group algebra space (Theorem 5.8).

Apologizing for my ignorance, but I have two questions that I cannot figure out straightforwardly following the derivations and arguments provided in the textbook:

(1) It can be shown (Theorem 5.5) that any permutation on a Young Tableaux gives an equivalent representation, so that $e_\lambda^p \equiv p e_\lambda p^{-1}$ is equivalent to $e_\lambda$. Since $e_\lambda^p$ is not necessarily a standard Young Tableaux, while it is an idempotent (since it is equivalent to an idempotent), the left ideal generated by it should be coincided with one of those generated by a standard Young Tableaux?

(2) Essentially primitive idempotent is defined (with more derivations given in the appendix IV of the textbook) as $e_\lambda \equiv s_\lambda a_\lambda$. I am wondering, by employing similar arguments used in the appendix (to be specific, one exchanges the line with column, symmetrizer with anti-symmetrizer in the arguments), it occurs to me that $\tilde{e}_\lambda \equiv a_\lambda s_\lambda$ is also essentially primitive idempotent. If I am wrong, which important piece I have missed, if not, will its corresponding left ideal be also coincided with some standard Young Tableaux ?

If I stray too far away from the right track, please delete this question. Many thanks.