Why is the hypoiodite anion unstable in solution, whereas hypochlorite is relatively stable?
When iodine is added to water, the following reaction results:

$$\ce{I2(l) + H2O(l) -&gt; OI^{-}(aq) + 2H+(aq) + I^{-}(aq)}$$

$\ce{I2}$ molecules and water molecules react to form hypoiodite ($\ce{OI-}$). The reaction can move both ways of the equilibrium, depending on the pH of the solution. Iodine may also occur as $\ce{I3-}$(aq), $\ce{HIO}$(aq), $\ce{HIO3}$(aq) . The compounds that are form behave differently when they come in contact with water. When iodine ends up in surface waters, it may escape as iodine gas.

$\ce{IO-}$ rapidly disproportionates to form $\ce{IO3-}$ and $\ce{I2}$
$$\ce{IO- -&gt; IO3- + I2}$$.

See here for more details. 

For the disproportionation reaction, consider the Latimer diagram for iodine:-



A species has a tendency to disproportionate into its two neighbor species if the potential on the right of species in a Latimer diagram is more positive than that on the left. In this case, right hand potential($\ce{+0.42}$) is greater than the left hand potential ($\ce{+0.15}$). So , it disproportionates.