Why does iodine solution turn transparent on adding base?
Iodine disproportionates, forming iodide and iodate ions, in the presence of base.  But it can be restored by neutralizing the base.  Neutralization will drop the hydroxide concentration low enough so that the disproportionation does reverse.

Let's punch in some numbers using the equilibrium constants given here.  We have

$$
\begin{array}{ccc}
\ce{I2 + 2OH- &amp;&lt;=&gt;&amp; I- + IO- + \ce{H_2O}}&amp;K=30\\
\ce{3IO- &amp;&lt;=&gt;&amp; 2I- + IO3-}&amp;K=10^{20}
\end{array}
$$

To superpose the reactions we take three units of the first one plus one unit of the second, canceling out the unstable hypoiodite ions.  So, rendering that three units of the first reaction mean three factors of that value of $K$, we end with:

$$\ce{3I2 + 6OH- &lt;=&gt; 5I- + IO3- + 3H2O}, K=2.7×10^{24}$$

If we consider a hypothetical case where all the iodine species are in their standard states, including iodine as the element, then the equilibrium hydroxide ion concentration is  $\pu{8.5×10^{-5} M}$.  Taking the base 10 logarithm and adding $\mathrm{p}K_\mathrm{w}=14$ then gives a $\mathrm{pH}$ just below $10$, so a neutral solution gets your elemental iodine back.