= Solution
Let $\lambda,\mu\vdash n$ each be a <partition of an integer>. Pad the lists of parts with zeros and use the <dominance order on partitions>
$$
\boxed{\lambda\unlhd\mu\quad\Longleftrightarrow\quad
\sum_{r=1}^h\lambda_r\leq\sum_{r=1}^h\mu_r\text{ for every }h\geq1.}
$$
The total sums agree, so only finitely many inequalities matter. This is a <partial order>: reflexivity and transitivity hold term by term, and equality of all partial sums forces equality of every part, proving antisymmetry. For example $(2,2)\unlhd(3,1)$. It compares concentration in the top rows, whereas diagram inclusion compares individual cells and is a different relation.
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