Can I electrolyze saturated sodium hydroxide?
At $T=25\ \mathrm{^\circ C}$, saturated sodium hydroxide solutions contain about $50\ \%$ sodium hydroxide. Hence, the concentration of $\ce{Na+}$ is about $c(\ce{Na+})=19\ \mathrm{mol\ l^{-1}}$ and the concentration of $\ce{H3O+}$ corresponds roughly to $\mathrm{pH\approx15}$, i.e. $c(\ce{H+})\approx10^{-15}\ \mathrm{mol\ l^{-1}}$ (neglecting activity coefficients).

The positive ions $\ce{Na+}$ and $\ce{H3O+}$ are attracted to the cathode. 
At the cathode, the redox potentials $E$ imply that, even under the given alkaline conditions, reduction of $\ce{H+}$ to $\ce{H2}$ is still preferred over reduction of $\ce{Na+}$ to $\ce{Na}$:

$$\begin{alignat}{3}
\ce{Na+ + e- \;&amp;&lt;=&gt; Na}\qquad &amp;&amp;E^\circ = -2.713\ \mathrm V\qquad &amp;&amp;E&amp;&amp;=E^\circ+0.05916\ \mathrm{V}\times\log_{10} [\ce{Na+}]\\
&amp; &amp;&amp; &amp;&amp; &amp;&amp;=-2.713\ \mathrm V+0.05916\ \mathrm{V}\times\log_{10} 19\\
&amp; &amp;&amp; &amp;&amp; &amp;&amp;=-2.64\ \mathrm V\\[6pt]
\ce{2H+ + 2e- \;&amp;&lt;=&gt; H2}\quad\quad &amp;&amp;E^\circ = +0.000\ \mathrm V\qquad &amp;&amp;E&amp;&amp;=E^\circ+0.05916\ \mathrm{V}\times\log_{10} [\ce{H+}]\\
&amp; &amp;&amp; &amp;&amp; &amp;&amp;=0.000\ \mathrm V+0.05916\ \mathrm{V}\times\log_{10}(10^{-15})\\
&amp; &amp;&amp; &amp;&amp; &amp;&amp;=-0.89\ \mathrm V
\end{alignat}$$

Therefore, the thermodynamically preferred reaction is the reduction of water to hydrogen according to
$$\ce{2H+ + 2e- -&gt; H2 }$$
or in this case better (in view of the given alkaline conditions)
$$\ce{2H2O + 2e- -&gt; H2 + 2OH- }$$

Nevertheless, under the given alkaline conditions, the concentration of $\ce{H3O+}$ is very low. The predominant positive ions are $\ce{Na+}$, which migrate to the cathode. Thus, especially when applying a high current density, the major part of the current of positive charges that reach the cathode is carried by $\ce{Na+}$. Therefore, $\ce{Na+}$ actually can be discharged at the cathode:
$$\ce{Na+ + e- -&gt; Na}$$
The ratio of discharged $\ce{Na+}$ to $\ce{H+}$ also depends on the cathode material. For some materials, the reduction of $\ce{H+}$ to $\ce{H2}$ is kinetically inhibited, which results in a significant overpotential.

However, any generated $\ce{Na}$ immediately reacts with water (except when a mercury cathode is used): 
$$\ce{2Na + 2H2O -&gt; 2Na+ + H2 + 2OH- }$$

Therefore, the net reaction is again the above-mentioned reduction of $\ce{H2O}$ to $\ce{H2}$. The $\ce{Na+}$ ions act as electrolyte and might be considered to act as a catalyst:

$$\begin{align}
\ce{2Na+ + 2e- &amp;-&gt; 2Na}\\
\ce{2Na + 2H2O &amp;-&gt; 2Na+ + H2 + 2OH- }\\
\hline\ce{2H2O + 2e- &amp;-&gt; H2 + 2OH- }
\end{align}
$$