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
$L\to M$