= Solution
Label the incoming <four-momenta> $p_1,p_2$ and the outgoing ones $p_3,p_4$. Define the <Mandelstam variables> $s=(p_1+p_2)^2$, $t=(p_1-p_3)^2$, $u=(p_1-p_4)^2$. Two incoming $\phi$ particles have <conserved charge> $+2$, so the only <tree-level Feynman diagram> is an $s$-channel $\Phi$ exchange, shown in panel (b) of <1/b>'s figure. Each end converts two charge-one lines into one charge-two line. There are no $t$- or $u$-channel exchanges: an incoming and an outgoing particle at such an end would transfer zero charge, for which this theory has no cubic <interaction vertex>.
$$
i\mathcal M_{\phi\phi}=(-ig)^2\frac{i}{s-M^2+i\epsilon},
\qquad
\boxed{\mathcal M_{\phi\phi}=-\frac{g^2}{s-M^2+i\epsilon}.}
$$
The single <Feynman diagram> is already symmetric under exchanging the two outgoing particles. This does not add another copy of the <scattering amplitude>; the <identical final-state symmetry factor> enters the integrated <scattering cross-section> instead.
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