= Solution
Applying a <permutation> to the elements of an $r$-element subset preserves its cardinality. The identity fixes each subset and
$$
\sigma\cdot(\tau\cdot U)=\{\sigma(\tau(u)):u\in U\}=(\sigma\tau)\cdot U,
$$
so this is the <symmetric group action on subsets>. For two $r$-element subsets $U,W$, choose a bijection $U\to W$ and a bijection between their complements; together they define a <permutation> of the full $n$-element set. Hence \b[the action is transitive].
A member of the <stabiliser subgroup> must permute $U$ internally and its complement internally, with independent choices. Thus
$$
\boxed{|(S_n)_U|=r!(n-r)!,\qquad |\operatorname{Orb}(U)|=\binom nr.}
$$
This also agrees with the <orbit-stabilizer theorem>; $0!=1$ handles $r=n$.
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