Let $X,Y$ be topological spaces. Two embeddings $f,g:X\rightarrow Y$ are equivalent if there is a homeomorphism $h:Y\rightarrow Y$ such that $h\circ f=g$.
Two subsets $A,B$ of $Y$ are equivalent if there is a homeomorphism $h:Y\rightarrow Y$ such that $h(A)=B$.
Clearly, if $f$ and $g$ are equivalent embeddings then the subsets $f(X)$ and $g(X)$ are equivalent. However, is the converse always true? That is, if $f$ and $g$ are embeddings such that $f(X)$ and $g(X)$ are equivalent, does it follow that $f$ and $g$ are equivalent embeddings?
Edit: Ambient isotopic embeddings are equivalent embeddings (although the reverse implication is not true. The trefoil knot embedding of $S^{1}$ in $3$-space and its mirror image are equivalent embeddings, but not ambient isotopic). However, the analog converse of the statement above for ambient isotopic subsets is not true. See the comments here.
Just to take this question off the unanswered list. Consider $Y={\mathbb R}$ with the standard topology and $X=\{0, 1, 2\}\subset Y$ with the subspace (discrete) topology. Take $f$ to be the identity embedding. Define $g: X\to Y$ by $g(0)=0, g(1)=2, g(2)=1$. Then, of course, $f(X)=g(X)$ are equivalent subsets, but $f$ is not equivalent to $g$.