Solution
= Solution
For type $(1^n)$, all labels are distinct, so a semistandard tableau is precisely a <standard Young tableau>. Thus $K_{\lambda,(1^n)}=f^\lambda$, and <Young's rule> gives
$$
\boxed{M^{(1^n)}\cong\bigoplus_{\lambda\vdash n}f^\lambda S^\lambda,\qquad
m_\lambda=f^\lambda=\frac{n!}{\prod_{c\in\lambda}h(c)}.}
$$
This is also the decomposition of the <regular representation>, since <tabloids> with one entry per row are naturally indexed by all <permutations>.