Hydroxylamine reduces $\mathrm{Fe}{}^{3+}$ and is itself oxidised to $\mathrm{N}{}_2\mathrm{O}$: $\mathrm{NH}{}_2\mathrm{OH}+\mathrm{Fe}{}^{3+}\to \mathrm{N}{}_2\mathrm{O}+\mathrm{Fe}{}^{2+}+\mathrm{H}{}^{+}+\mathrm{H}{}_2\mathrm{O}$. The n-factor of $\mathrm{NH}{}_2\mathrm{OH}$ in this reaction is

N in $\mathrm{NH}{}_2\mathrm{OH}$ is -1 and in $\mathrm{N}{}_2\mathrm{O}$ it is +1, so each N loses 2 electrons, and the n-factor per $\mathrm{NH}{}_2\mathrm{OH}$ is 2. The balanced equation is $2\mathrm{NH}{}_2\mathrm{OH}+4\mathrm{Fe}{}^{3+}\to \mathrm{N}{}_2\mathrm{O}+4\mathrm{Fe}{}^{2+}+4\mathrm{H}{}^{+}+\mathrm{H}{}_2\mathrm{O}$; 4 electrons are transferred for 2 $\mathrm{NH}{}_2\mathrm{OH}$, which also gives 2 per molecule.