# Do two embeddings of a Euclidean space into a higher dimensional one only differ by a diffeomorphism?

Let $d\le n$ and $$f,g\colon\mathbb{R}^d\hookrightarrow\mathbb{R}^n$$ be two smooth embeddings. Is there a diffeomorphism $$\phi\colon\mathbb{R}^n\rightarrow \mathbb{R}^n,$$ such that $$f=\phi\circ g$$ holds?

In other words, does the diffeomorpshism group $\operatorname{Diff}{(\mathbb{R}^n)}$ act transitively on the set of embeddings $\operatorname{Emb}(\mathbb{R}^d,\mathbb{R}^n)$ by postcomposition?

As the other answer suggests, there is an obstruction in the form of properness. If $f$ is a closed (synonymously, proper) embedding, then so is $\phi \circ f$. But even restricting to closed embeddings this is not true.
Consider the case of long knots. That is, embeddings of $\mathbb R^1 \hookrightarrow \mathbb R^3$ that are the standard embedding outside $[-n,n]$ for some $n$. Then any two long knots can be mapped from one to the other by a diffeomorphism if and only if they are (compactly supported) isotopic. In other words, the same obstructions as in knot theory - not all knots $K,L: S^1 \hookrightarrow S^3$ have a diffeomorphism taking $K$ to $L$ (for instance, the trefoil and the unknot), the same is true for long knots.
This same obstruction is going to be true for long embeddings $\mathbb R^{n-2} \hookrightarrow \mathbb R^n$; every long embedding $\mathbb R^{n-1} \hookrightarrow \mathbb R^n$ is diffeomorphic to the standard one (except possibly for $n=4$? but in this case we still have that they're PL homeomorphic) by the Schoenflies theorem; all other long embeddings $\mathbb R^{n-k} \hookrightarrow \mathbb R^n$ are PL homeomorphic to the standard one but not necessarily diffeomorphic.
To avoid dealing with long knots and proper embeddings or whatever you might prefer to work with smooth embeddings $S^k \hookrightarrow S^n$. This removes technical complications. See Levine for the classification of high-codimension smooth knots. See Zeeman for the proof that all high-codimension knots are PL-standard.
If $g$ is onto and $f$ is not that cannot be possible. That can't be either if $d < n$.