The reaction of $\mathrm{KMnO}_4$ and potassium iodide ($\mathrm{KI}$) in acidic medium can be represented as follows:



In this reaction, $\mathrm{KMnO}_4$ acts as an oxidizing agent, while $\mathrm{KI}$ acts as a reducing agent. The oxidation state of manganese changes from +7 in $\mathrm{KMnO_4}$ to +2 in $\mathrm{Mn^{2+}}$, which is a reduction of 5 units.

Therefore, the total change in the oxidation state of manganese involved in the reaction is $\boxed{5}$.