= Solution
For shape and type both $\mu$, column strictness makes every entry in row $r$ at least $r$. All $\mu_1$ entries equal to one must therefore be in the first row, which has exactly that length; it is filled with ones. Remove that row and repeat after subtracting one from the remaining labels. <Induction> gives the unique filling whose $i$th row consists of $i$'s. Thus $K_{\mu\mu}=1$, and <Young's rule> proves that $S^\mu$ occurs exactly once.
More generally, if a semistandard filling of shape $\lambda$ and type $\mu$ exists, its entries at most $k$ must lie in the first $k$ rows. Counting them gives
$$
\sum_{i=1}^k\mu_i\le\sum_{i=1}^k\lambda_i\qquad\text{for every }k.
$$
These are the inequalities defining the <dominance order on partitions>. Hence $K_{\lambda\mu}>0$ implies $\boxed{\lambda\ge\mu}$, and the asserted restriction on summands follows from <Young's rule>.
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