# What does invariance of Lagrangian under a group action mean?

Let $$L(q_i,\dot{q_i},t)$$ be the(a?) Lagrangian of a physical system. Assume that the gen. coordinates $$q_i$$ transform under a certain Group G as $$q_i\rightarrow q_i'=f_i(q_j,\theta_k)$$ where $$f_i$$ are functions depending upon $$q_i$$'s and parameters $$\theta_k$$'s satisfying appropriate conditions (invertibility etc.)

Does the Lagrangian transform as $$L\rightarrow L'=L(q_i',\dot{q_i}',t)$$ or as $$L\rightarrow L'=L(q_i(q_i'),\dot{q_i}(q_i',\dot{q_i}'),t)$$ or in some other way?

What does it mean when we say that $$L$$ is invariant under transformations of group G? Do we mean that the numerical value of $$L$$ is unchanged at any t? Or the form of $$L'(q_i',\dot{q_i}',t)$$ is same as that of $$L(q_i,\dot{q_i},t)$$? Or something else?

For example, let's take the Lagrangian density of complex scalar field (I seem to be more comfortable with field theory formulation) - $$\mathcal{L}=\left(\partial_{\mu} \Phi\right)^{\dagger} \partial^{\mu} \Phi-\left(\Phi^{\dagger} \Phi\right)$$ under group G=$$U(1)$$, \begin{aligned} \Phi \rightarrow \Phi^{\prime} &=e^{-i \alpha} \Phi \\ \Phi^{\dagger} \rightarrow \Phi^{\prime \dagger} &=e^{i \alpha} \Phi^{\dagger} \end{aligned} What would be the expression of transformed $$\mathcal{L\rightarrow L}'$$=?

Again, what does "invariance of $$\mathcal{L}$$ under $$U(1)$$" mean, both physically and mathematically?