KIO₃ + 5KI + 6HCl → 3I₂ + 6KCl + 3H₂O Correct statement of this reaction is

${\mathrm{KIO}}_3+5 \mathrm{K}\mathrm{I}+6\mathrm{HCl}\to 3{\mathrm{I}}_2^0+6\mathrm{KCl}+3{\mathrm{H}}_2\mathrm{O}$

KIO₃ (Oxidizing agent) reduces to I₂
KI  (reducing agent)  oxidizes to I₂ 

This is a comproportion reaction Oxidation number of I change from +5 to zero and -1 to zero