= Solution
Label the incoming <fermion> <momenta> $p_1,p_2$ and outgoing <fermion> <momenta> $p_3,p_4$, with $p_1+p_2=p_3+p_4$. There are two <tree-level Feynman diagrams>: the $t$ channel connects $1$ to $3$ and $2$ to $4$, with exchanged <momentum> $q_t=p_1-p_3=p_4-p_2$; the $u$ channel connects $1$ to $4$ and $2$ to $3$, with $q_u=p_1-p_4=p_3-p_2$. Both internal lines are <auxiliary-field propagators>. The upper row in the figure gives these labelled connections; the outgoing labels are exchanged in its second panel to avoid a visually unnecessary crossing.
Each <tree-level Feynman diagram> contains two factors $ig$ and one factor $-i$, giving $ig^2$. The exchange of the two identical outgoing <fermions> gives the relative <fermionic sign>. Up to a common external-state phase, the corresponding <scattering amplitude> is
$$
\boxed{\mathcal M_{ff}=g^2\left[(\bar u_3u_1)(\bar u_4u_2)-(\bar u_4u_1)(\bar u_3u_2)\right].}
$$
There is no annihilation channel: both incoming particles carry the same <fermion number>, whereas a $\sigma\bar\psi\psi$ <interaction vertex> couples one <fermion> to one <antiparticle> when both are incoming.
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