I have a smooth (possibly compact, or closed, or oriented, or more than one of the previous) $n$-manifold $M$ together with a fixed volume form $\rho\in\Omega^n(M)$. Can $M$ be embedded into some Euclidean space in such a way that the volume form is sent to the restriction of the standard Euclidean volume form to $M$?
Formally, I am looking for an embedding $i:M\to\mathbb{R}^N$ (for some integer $N$) such that $$i^*d\operatorname{vol}_{\mathbb{R}^N} = \rho,$$ where $d\operatorname{vol}_{\mathbb{R}^N}=dx^1\wedge\ldots\wedge dx^N$ is the usual volume form.
A positive answer would follow from Nash's embedding theorem provided we can equip $M$ with a Riemannian metric inducing $\rho$ as volume form.
Let $M$ be a smooth $n$-manifold with a fixed volume form $\rho$. Equip it with a Riemannian metric $g$ (which exists for example by taking the pullback of the standard metric on Euclidean space after embedding $M$ by Whitney's theorem), and denote by $*_g$ the Hodge-$*$ operator with respect to $g$. If $\mu\in C^\infty(M)$ is everywhere strictly positive we have $$*_{\mu g} = \mu^{\frac{n}{2}-p}*_g$$ on $p$-forms. The volume form induced by $g$ is given by $*_g1$, thus we look at the equation $$1 = *_{\mu g}\rho=\mu^{-\frac{n}{2}}*_g\rho.$$ This is solved by setting $\mu = (*_g\rho)^{\frac{2}{n}}$, showing that in fact every conformal class on $M$ has a representative inducing the desired volume form as its volume form.