In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of ${\mathrm{CO}}_2$ is:

$2\mathrm{ }{\mathrm{MnO}}_4^{-}+5{\mathrm{C}}_2{\mathrm{O}}_4^{2-}+16{\mathrm{H}}^{+}\to 10\mathrm{ }{\mathrm{CO}}_2+2\mathrm{ }{\mathrm{Mn}}^{2+}+8{\mathrm{H}}_2\mathrm{O}$

$\mathrm{C}$ goes from $+3$ oxidation state to $+4$ oxidation state.

Total $10 {\mathrm{e}}^{-}$ are transferred for $10$ molecules of ${\mathrm{CO}}_2$.

Hence, for $1$ molecule of ${\mathrm{CO}}_2$, only one ${\mathrm{e}}^{-}$ is transferred.
So, number of electrons involved for $1$ molecule of ${\mathrm{CO}}_2$  is $1$.