OP's underlying main question seems to be the fact that (if we are using the standard convention where each term$^1$ in the Lagrangian is divided by its symmetry factor) then the numerical coefficient in front of the Feynman diagram is the reciprocal of its symmetry factor $S$. For a proof, see e.g. Ref. 1.

References:

1. P. Etingof, Geometry & QFT, MIT 2002 online lecture notes; Chapter 3.

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$^1$ In particular, this means that OPs interaction term in the Lagrangian should be normalized as $-\frac{\lambda}{4}\phi^{\ast 2}\phi^2$.

OP's underlying main question seems to be the fact that (if we are using the standard convention where each term$^1$ in the Lagrangian is divided by its symmetry factor) then the numerical coefficient in front of the Feynman diagram is the reciprocal of its symmetry factor $S$. For a proof, see e.g. Ref. 1.

References:

1. P. Etingof, Geometry & QFT, MIT 2002 online lecture notes; Chapter 3.

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$^1$ In particular, this means that OP's interaction term in the Lagrangian should be normalized as $-\frac{\lambda}{4}\phi^{\ast 2}\phi^2$.