Invariant vector fields and forms- should I have paid more attention in Linear algebra? My question regards invariant vector fields and differential forms on Lie groups, but I have a feeling it may just boil down to the well known fact that the dual vectorspace is not canonically isomorphic to the given one. 
Notation: Let $L$ be a Lie group, and $e \in L$ the identity element. We can take a vector $V_e\in T_eL$ and using Lie group structure produce a vector field $V$ on $L$, where 
$$
V_g := g_* V_e.
$$
Similarly, we can take a covector $\phi_e \in T^*_e L$ and produce a differential form $\phi$ on $L$, where
$$
\phi_g := (g^{-1})^* \phi.
$$
Question: Using the above notation, suppose we start with just $L$ and a vector $V_e$ at the identity. I'm wondering if the following two constructions produce the same differential form on $L$:


*

*Take the covector $V_e^*$, the dual to $V_e$, and use that to produce a differential form $\phi$ on $L$.  

*Use $V_e$ to produce a vector field $V$ on $L$, and then take the dual differential form $V^*$.
Does 
$$ \phi = V^*?$$
Attempt: 


*

*Invariant differential forms are determined by their value at the identity, so since $\phi$ and $V^*$ are clearly equal at $e$, if $V^*$ is invariant then $\phi = V^*$. I'm not sure how to prove that though.

*Actually, $\phi$ and $V^*$ are equal on all of $V$. But this isn't enough to directly say they are equal. (e.g. $dx$ and $dx + dy$ along the vector $\frac{\partial}{\partial x}$.)

*Maybe what's going on is that since only a "1- dimensional" amount of information has been given and identifying a space with its dual is non-canonical, I only know about 1 dimension of these forms and they happen to agree in that dimension by construction. 
EDIT: I've written a follow-up question which actually makes sense!
 A: Your third point is the most relevant to your question. In linear algebra, given a vector $v$ in a finite dimensional space $V$ without any additional structure, there is no natural notion of a "dual vector" in $V^{*}$, only a dual basis. That is, only if you choose a whole basis $(v_1,\dots,v_n)$, you can use it to define a dual basis $(v^1, \dots, v^n)$ of $V^{*}$ (that satisfies $v^i(v_j) = \delta^i_j$). 
You might think that this is mostly nick picking and say "Given a vector $0 \neq v \in V$, we can complete it in an arbitrary way to a basis $\mathcal{B} = (v_1 = v, v_2, \dots, v_n)$ of $V$, take the dual basis $\mathcal{B}^{*} = (v^1, \dots, v^n)$ and define the dual vector of $v$ to be $v^1$". Unfortunately, the resulting vector $v^1$ depends not only on $v_1$ but on the whole sequence $(v_1,\dots,v_n)$. Different choices of bases extending $v$ will result in different dual covectors.
Thus, unless you specify more information, your question doesn't make sense as it is because you don't have a procedure to construct a dual covector to a single vector.
