For an aqueous solution of $\mathrm{Al}{}_2(\mathrm{SO}{}_4){}_3$, the equivalent conductivity ${\Lambda }_{eq}$ is related to the molar conductivity ${\Lambda }_m$ by

${\Lambda }_{eq}={\Lambda }_m/n$, where $n$ is the total charge carried by the cations (or anions) per formula unit. In $\mathrm{Al}{}_2(\mathrm{SO}{}_4){}_3$, two $\mathrm{Al}{}^{3+}$ carry $2\times 3=6$ positive charges (three $\mathrm{SO}{}_4{}^{2-}$ carry 6 negative charges), so $n=6$ and ${\Lambda }_{eq}={\Lambda }_m/6$. The count of ions, 5, is not the n-factor.