Let $X$ be a real $n$ dimensional manifold. One knows that it can be embedded into $\mathbb{R}^{2n}$ by the Whitney embedding theorem. The normal bundle for such an embedding will be a rank $n$ real vector bundle. I would be interested in understanding the connection between this embedding, its normal bundle and the rank $n$ bundles on $X$.

How does the normal bundle change under the isotopy of this embedding?

Is it possible to obtain all the isomorphism classes of rank $n$ vector bundles by such isotopies?

If the answer for 2. is negative, is it still possible to find an embedding of $X$ into $\mathbb{R}^{2n}$ for each vector bundle of rank $n$, such that the corresponding normal bundle is isomorphic to it?

I was thinking that the answer should have something to do with the self-intersection of $X$ in the vector bundles and its self-intersection in $\mathbb{R}^{2n}$, but don't really see how to use that.

**Edit:** I have realized that it shouldn't be possible to obtain a vector bundle with a non-zero self intersection of the zero section like this. However, I would still like to know if it works for all the other cases.

**Edit 2:** Based on Mike's comment, I have realized I have missed something very obvious. I am just thinking whether one still gets all the rank $n$ vector bundles which complete the tangent bundle to a trivial $2n$ bundle.

any$N$ have the property that $TM \oplus \nu$ is trivial. That is, $\nu$ is a "stable additive inverse" of $TM$. This completely determines its Stiefel-Whitney classes from those of $TM$ and thus is hugely constrained. $\endgroup$ – Mike Miller Jul 27 '19 at 19:53