Is the dimension given by Klee trick ever sharp?

Question

The Klee Trick allows one to find an $\mathbb{R}^m$ where two embeddings of same compact metric space have homeomorphic complements. More precisely, given two embeddings of a compact metric space $K$ into $\mathbb{R}^n$, $f_{n,1}$, $f_{n,2}$, we can construct two embeddings of $K$ into $\mathbb{R}^{2n}$ such that the images of $K$ are equivalent under a homeomorphism of $\mathbb{R}^{2n}$. (The trick itself produces an isotopy of the the two embeddings in $\mathbb{R}^{2n}$.)

However, I was wondering if there is an example of a compact metric space $K$ and a pair of embeddings $f_{n,1},f_{n,2}:K \rightarrow \mathbb{R}^n$ such that the embeddings are not equivalent under a homeomorphism of $\mathbb{R}^{n+m}=\mathbb{R}^n \times\mathbb{R}^m$ for all $m < n$.

To clarify with a non-example, any two (tamely) embedded knots in $\mathbb{R}^3$ will be isotopic in $\mathbb{R}^4=\mathbb{R}^3 \times\mathbb{R}$, and so for embeddings of $S^1$ into $\mathbb{R}^3$, the isotopy obtained from the Klee trick is not optimal.

Answer 1

Are the embeddings required to be isometric embeddings? If not, then what about including three points into $\mathbb{R}$ in two ways, so that the middle point of the three changes? More explicitly, let $K=\{a,b,c\}$ and define $f(a)=0$, $f(b)=1$, $f(c)=2$, whereas $g(a)=1$, $g(b)=0$, $g(c)=2$. There isn't any self-homeomorphism of $\mathbb{R}$ whose composition with $f$ is equal to $g$.