In $\mathrm{MnO}_{4}^{-}$ an electron is momentarily changing $\mathrm{O}^{--}$ to $\mathrm{O}^{-}$ and reducing the oxidation state of the metal from $Mn\,(VII)$ to $Mn\, (VI)$. Charge transfer requires that the energy levels on the two different atoms are fairly close.

$\,\begin{array}{*{20}{c}}
  {O\, = \,(8)\, = 2,} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,K} 
\end{array}\,\begin{array}{*{20}{c}}
  {6\,;} \\ 
  L 
\end{array}$           $\,\begin{array}{*{20}{c}}
  {Mn\,\,(25)\, = 2,} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,K} 
\end{array}\,\begin{array}{*{20}{c}}
  {8,} \\ 
  L 
\end{array}\begin{array}{*{20}{c}}
  {15} \\ 
  M 
\end{array}$

Hence the charge transfer occurs from

$L\to M$