About Feynman rules and symmetry factor

Question

I have two simple questions about Feynman rules for Lagrangian that have a derivative of fields.

1. For example for Lagrangian in this link part a with derivative couplings. Is the vertex depend on direction of momentum? If we have a vertex and then change the momentum direction does it change? (depend on incoming and outgoing direction.)

2. And other simple question is that the symmetry factor of diagram is invariant under the change of interaction term and only depends on geometry of the diagram, is it right?for example if we change $1/4!$ of $\phi^4$ interaction to $1$, the same diagram have a same symmetry factor?

Answer 1

1. Yes, the Feynman rule (in momentum space) for interaction vertices with derivatives may in general depend on the orientation of the $4$-momentum of its legs. (If there is an even number of derivatives, the orientation doesn't matter.) Be aware that different consistent conventions exist in the literature.

2. Yes, the symmetry factor of a Feynman diagram does by definition not depend on the normalization of the coupling constant. On the other hand, (the value of) a Feynman diagram will in general obviously depend on the coupling constant and its normalization. Concerning vertex factors, see also this related Phys.SE post.