Should the sign of Feynman rule factors be the same as the sign in the Lagrangian?

Suppose I have this Lagrangian

$$L = \frac{1}{2}\partial^\mu\phi\partial_\mu\phi - \frac{1}{2}m^2\phi^2-\frac{g_3}{3!}\phi^3$$

One of the Feynman rules would be associating a factor $$(-ig_3)$$ to each $$\phi^3$$ vertex. This could be seen from the evaluation of the two-point function $$\langle0|T\{\phi(x)\phi(y)\}|0\rangle$$. My question is if we replace '-' with '+' in $$\phi^3$$ term, should we flip the sign of the factor also? Thus the sign of Feynman rule factors is the same as the sign of original terms in the Lagrangian?

If we add counterterms to this Lagrangian,

$$L= \frac{1}{2}\partial^\mu\phi\partial_\mu\phi - \frac{1}{2}m^2\phi^2-\frac{g_3}{3!}\phi^3 +\frac{1}{2}(\partial^\mu\phi\partial_\mu\phi)\delta_z-\frac{1}{2}\delta_m^2\phi^2\color{red}{-\delta_1\phi}-\delta_3\frac{\phi^3}{3!}.$$

There is a term (marked in red) we need to include, but it's not in the original Lagrangian. Why do we add $$-\delta_1\phi$$, rather than $$+\delta_1\phi$$?

Yes, the sign of the vertex Feynman rule $$\frac{i}{\hbar}(-g_n)$$ is directly correlated with the sign of the corresponding interaction term $$-\frac{g_n}{n!}\phi^n$$ in the action.