Invariant forms on a Lie group - are these equal?

Motivation: This is a follow-up question to this question that I asked a few minutes ago. (my question there actually doesn't even make sense (apparently I did not pay close enough attention) - see levap's answer. I've tried to modify that question so that it actually does make sense and here is what I've come up with:

Background: Let $L$ be a Lie group, and $g$ a left-invariant riemannian metric on $L$. With respect to $g$, choose an orthonormal basis $v_e = v_e^1, \cdots v_e^n$ for the tangent space at the identity $T_e L$, and let $w_e = w_e^1, \cdots, w_e^n$ denote the dual basis to $v_e$ in $T^*_eL$.

As $g$ is left-invariant this orthonormal basis at the identity induces a global an orthonormal frame field $v$ on all of $L$ by using Lie group structure. See here for details.

Now, we can construct global 1- forms from this basis in two obvious ways, and I'm wondering if the result is the same.

1. Take $w_e$, the dual basis to $v_e$, and using the Lie group structure define a global basis of invariant 1-forms. Again, see here for details. Call this $\phi =\phi^1, \cdots, \phi^n$.

2. Take the global frame field $v$ and at each point $p \in L$ take the dual coframe to produce a global coframe field. Call this $w= w^1, \cdots, w^n$.

Question: Does $$\phi = w?$$ i.e does $\phi^1 = w^1, \cdots, \phi^n = w^n$?

Attempt: I think I may have figured this out. We have by construction that at any point $g\in L$ $$w_g^i (v_g^j) = \delta_{ij}.$$ So for equality of $\phi$ and $w$ we would need $$\phi_g^i (v_g^j) = \delta_{ij}.$$ However, $$\phi_g^i (v_g^j)= (w_e^i \circ dg^{-1}) \circ (dg \circ v_e^j) = w_e^i(v_e^j) = \delta_{ij}.$$ So by uniqueness of the dual basis $\phi$ and $w$ are equal.

$(L_{g^{-1}})^*(w^i)(v^j_g)=w^i((L_{g^{-1}})_*v^j_g) = w^i(v^j_{g^{-1}g}) = w^i(v^j_e) = \delta^{ij}$ so yes, they're equal.