What is the number of electrons being transferred in electrolysis of aluminum oxide?
I guess the misunderstanding arises from the fact that you treat $n$ as the same entity as the lowest common multiple (LCM) used to balance the net equation using half-reactions for the electrolysis of molten aluminium(III) oxide:
$$
\begin{align}
&amp;\text{cathode:} &amp;\ce{Al^3+(melt) + 3 e- &amp;-&gt; Al^0(l)} &amp;|\cdot4\tag{red}\\
&amp;\text{anode:} &amp;\ce{2 O^2-(melt) &amp;-&gt; O2^0(g) + 4 e-} &amp;|\cdot3\tag{ox}\\
\hline
&amp; &amp;\ce{4 Al^3+ + 6 O^2- &amp;-&gt; 4 Al^0 + 3 O2^0}\tag{redox}
\end{align}
$$
or, finally, with the smallest whole-number coefficients:
$$\ce{2 Al2O3(l) -&gt; 4 Al(l) + 3 O2(g)}\label{rxn:R1}\tag{R1}$$
as well as normalized to a single formula unit:
$$\ce{Al2O3(l) -&gt; 2 Al(l) + 3/2 O2(g)}\label{rxn:R2}\tag{R2}$$
In the equation
$$Δ_\mathrm rG = -nFE$$
$n$ is the number that matches the amount of electrons (in mol — since we are using molar convention for Gibbs free energy) transferred in the balanced equation from a reducing agent to an oxidizing agent.
From the balanced reaction \eqref{rxn:R2} it is evident that the number of electrons $n$ transferred from $\ce{O^2-}$ to $\ce{Al^3+}$ is $3 \cdot 2 = 6,$ where $3$ is the number of oxygens per formula unit, and $2$ is the number of electrons each oxygen donates.