= Solution
A representation of the <one-arrow quiver> is an $n_2\times n_1$ matrix $F$. If $r=\operatorname{rank}F$, changes of basis put it in the form with an $r\times r$ identity block and all other entries zero. Since multiplication by invertible matrices preserves <matrix rank>,
$$
\boxed{\mathcal O_X=\{M\in\operatorname{Mat}_{n_2\times n_1}(k):\operatorname{rank}M=r\}.}
$$
This describes the orbit for every $0\leq r\leq\min(n_1,n_2)$, including the zero orbit. Its dimension is
$$
\boxed{\dim\mathcal O_X=r(n_1+n_2-r).}
$$
Indeed, choosing its $r$-dimensional image contributes $r(n_2-r)$ parameters in a <Grassmannian>, and choosing a surjective map from $k^{n_1}$ onto that image contributes $rn_1$. This is a <rank orbit of a matrix under left-right multiplication>.
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