Understanding sodium hydroxide drain cleaner's exothermic dissolution of aluminum; does the aluminum need to be oxidized before it dissolves?
If you want to understand the reaction of $\ce{NaOH}$ on Aluminium, you may describe it this way. First, the aluminium reacts spontaneously with water, like sodium or calcium, producing gaseous Hydrogen $\ce{H2}$ and an metallic oxide or hydroxide. With sodium or calcium, the hydroxide produced by this reaction quits the metal surface, and get dissolved, leaving a fresh surface of metal ready to react with more water. The reaction with sodium is: 

$$\ce{2 Na + 2 H2O -&gt;   2 NaOH  +  H2}$$ 

With aluminium, it is not the case. The reaction is:

$$\ce{4Al + 6 H2O -&gt; 2 Al2O3 + 6 H2}$$

But the aluminium oxide is not soluble into water and it makes a waterproof layer on the surface of the metal, which prevents further contact between the metal and water. So the reaction stops, due to this protecting effect of $\ce{Al2O3}$.

There are two ways of allowing the reaction to continue: 


Dissolve $\ce{Al2O3}$ as soon as it is produced by adding $\ce{NaOH}$, as @Uhoh tells it in its first equation. 
Prevent the $\ce{Al2O3}$ oxide layer from adhering on the metal surface by adding some mercury salt in the water. In this case the mercury ions $\ce{Hg^{2+}}$ may go through the oxide layer, reacts with the metal and produces a mercury deposit which forms an alloy with Aluminium. The reaction is: 


$$\ce{2Al + 3 Hg^{2+} -&gt; 2 Al^{3+} + 3Hg}$$

The $\ce{Al2O3}$ layer cannot adhere on the surface of this alloy Al-Hg. So the 
 reaction with water continues as long as there is some mercury atoms left on the 
 metallic surface. 

So, to go back to @Uhoh's question, it is not $\ce{NaOH}$ that acts on Aluminium. It is water. The action of $\ce{NaOH}$ is to remove the oxide layer.