6 NaI (aq) + 2 KMnO4 (aq) + 4 H2O (l) → 3 I2 (s) + 2 MnO2 (s) + 6 NaOH (aq) + 2 KOH (aq)
This is an oxidation-reduction (redox) reaction:
2 MnVII + 6 e- → 2 MnIV (reduction)
6 I-I - 6 e- → 6 I0 (oxidation)
KMnO4 is an oxidizing agent, NaI is a reducing agent.