The action of iodine is part of a long reaction chain with chlorate, so not a direct action as implied.
I start with iodine and water equilibrium:
$\ce{I2 + H2O <=> HI + HOI }$
Next more complex part of the reaction sequence is discussed here, to quote:
Chlorate ion oxidizes iodide ion to hypoiodous acid and chlorous acid in the slow and rate-determining step:
$\ce{ClO3− + I− + 2 H+ → HOI + HClO2}$
Chlorate consumption is accelerated by reaction of hypoiodous acid to iodous acid and more chlorous acid:
$\ce{ClO3− + HOI + H+ → HIO2 + HClO2}$
More autocatalysis when newly generated iodous acid also converts chlorate in the fastest reaction step:
$\ce{ClO3− + HIO2 → IO3− + HClO2}$
Chlorous acid is unstable, disproportionating to hypochlorous acid and chloric acid:
$\ce{2 HClO2 → HOCl + H+ + ClO3- }$
Also, the known action of hypochlorous acid (even H2O2) on HI liberating iodine:
$\ce{2 HOCl + 2 HI -> 2 HCl + I2}$
Finally, some chlorine evolution:
$\ce{HCl + HOCl <=> Cl2 + H2O }$
Further reference on the complexity of the reaction system, see Page 94, Table 5.2 in this ebook.