$2{\mathrm{Cu}}_2{\mathrm{O}}_{\left(\mathrm{s}\right)}+{\mathrm{Cu}}_2{ \mathrm{S}}_{\left(\mathrm{s}\right)}\to 6{\mathrm{Cu}}_{\left(\mathrm{s}\right)}+{\mathrm{SO}}_{2( \mathrm{g})}$ the oxidant and reductant respectively in the above reaction are

The oxidation states are as: 

$\stackrel{+1}{{\mathrm{Cu}}_2\mathrm{O}\left(\mathrm{S}\right)}+{\mathrm{Cul}}_2\mathrm{S}\left(\mathrm{S}\right)\to 6{\mathrm{Cu}}_{\left(\mathrm{S}\right)}+{\mathrm{SO}}_{2( \mathrm{g})}^{+4-2}$

Copper is reduced from +1 state to zero. 

Sulphur is oxidised from -2 to -4. 

${\mathrm{Cu}}_2\mathrm{O}$helps sulphur in ${\mathrm{Cu}}_2 \mathrm{S}$ to increase its oxidation number, hence $\mathrm{Cu}$ $\left(\mathrm{I}\right)$ is an oxidant and sulphur  of ${\mathrm{Cu}}_2 \mathrm{S}$ helps copper both in ${\mathrm{Cu}}_2 \mathrm{S}$ and ${\mathrm{Cu}}_2\mathrm{O}$ to decrease its oxidation number, hence $\mathrm{S}$ of ${\mathrm{Cu}}_2 \mathrm{S}$ is a reductant. And the given reaction is a redox reaction