To solve this problem, we need to analyze the redox reaction involving $\mathrm{KMnO_4}$ and iodide ions in a neutral or faintly alkaline medium.$\\$In this reaction, $\mathrm{KMnO_4}$ acts as an oxidizing agent and oxidizes iodide ions $(\mathrm{I^-})$ to iodate ions $(\mathrm{IO_3^-}). \\$The oxidation state of manganese in $\mathrm{KMnO_4}$ is $+7. \\$The balanced half-reaction for the reduction of $\mathrm{KMnO_4}$ in a neutral or faintly alkaline medium is:$\\ \mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-} \\$In this half-reaction, manganese is reduced from an oxidation state of $+7$ in $\mathrm{MnO_4^-}$ to $+4$ in $\mathrm{MnO_2}. \\$Thus, the change in oxidation state of manganese is from $+7$ to $+4. \\$Therefore, the correct option is Option 1: $+7$ to $+4.$