Determine 1-form from volume forms Given a 1-form $\omega\in \Omega^1(\mathbb{R}^p)$ we can construct various non-trivial $p$-forms ($\omega\wedge\star\omega$ excluded), but using the exterior derivative for all of all the players to be wedged together once at most. Let's call this game "Reach the $p$-form".
For $p=2$ we can construct 
\begin{eqnarray}
s_1=&d\star\omega\\
s_2=&d\omega
\end{eqnarray}
For $p=3$
\begin{eqnarray}
t_1=&d\star\omega\\
t_2=&\omega\wedge d\omega\\
t_3=&\star\omega\wedge\star d\omega\\
t_4=&d\omega\wedge\star d\omega\\
\end{eqnarray}
How many of these volume forms can be build for each $p$? And the follow up question, which is more interesting, knowing all the volume forms, is it possible to reconstruct the initial 1-form?
A quick visualization of this can be seen below:
.
$\textbf{Edit 1}:$ The volume form $\omega\wedge\star\omega=<\omega,\omega> vol=1\,  vol$. I should have mention that.
$\textbf{Edit 2}$: Would it be possible to answer this question if we restrict it to non-constant coefficients?
$\textbf{Edit 3}$: @Robert Bryant Exactly, the norm is $1$ under the standard flat metric on  $\mathbb{R}^n$ and I want to assume that the $1$-form has non-constant coefficients.
 A: Here is a note to explain why the answer to Question 2 is still 'no', even if one restricts to $1$-forms of norm $1$ with non-constant coefficients.
Let $a$ and $b$ be (real-valued) functions on the plane $\mathbb{R}^2$ and consider the $1$-forms
$$
\alpha = \cos a\,\mathrm{d}x - \sin a\,\mathrm{d}y
\quad\text{and}\quad
\beta = \cos b\,\mathrm{d}x - \sin b\,\mathrm{d}y.
$$
Then 
$$
\star\alpha = \sin a\,\mathrm{d}x + \cos a\,\mathrm{d}y
\quad\text{and}\quad
{\star}\beta = \sin b\,\mathrm{d}x + \cos b\,\mathrm{d}y,
$$
so $\alpha\wedge\star\alpha = \beta\wedge\star\beta = \mathrm{d}x\wedge\mathrm{d}y$.
Now consider the two equations $\mathrm{d}\alpha = \mathrm{d}\beta$ and $\mathrm{d}\,{\star}\alpha = \mathrm{d}\,{\star}\beta$.  This is system of two partial differential equations for the two functions $a$ and $b$, and it is easy to see that, as long as $\sin(a-b)\not=0$, this is an elliptic determined system.  Thus, one can find solutions with $\alpha\not=\beta$ (at least locally), by specifying $a$ and $b$ as analytic functions along the $x$-axis that do not differ by an integer multiple of $\pi$. 
Thus, even setting aside the 'trivial' cases, you cannot always recover $\omega$ from knowing $\mathrm{d}\omega$ and $\mathrm{d}\,{\star}\omega$.
