NaIO3 (aq) + 5 NaI (aq) + 3 H2O (l) → 3 I2 (s) + 6 NaOH (aq)
This is an oxidation-reduction (redox) reaction:
5 I-I - 5 e- → 5 I0 (oxidation)
IV + 5 e- → I0 (reduction)
NaI is a reducing agent, NaIO3 is an oxidizing agent (synproportionation (comproportionation)).