Definition of ambiant isotopy that I used is : two knots K and L are ambiant isotopy if there exists a function $H:X \times I \to X$ such that $H$ is continuous and $H_t:X \to X$ is homeomorphism and $H_0(x)=x, H(K,1)=L$.
Definition of homeomorphism equivalent that I used is : two knots $K$ and $L$ are homeomorphic if there exists a homeomorphism $h:X \to X$ such that $h(K)=L$.
So why if k and L are homemorphism equivalent but not ambiant isotopy?
On
Suppose $X$ is the torus $X = S^1 \times S^1$, choose $p \in S^1$, let $K = \{p\} \times S^1$, and if $L = S^1 \times \{p\}$. There does indeed exist a homeomorphism $f : X \to X$ such that $f(K)=L$.
However, no such homeomorphism $f$ is isotopic to the identity, and in fact $f$ is not even homotopic to the identity. To see why, consider the first homology group, with its direct factorization arising from the Künneth formula: $$H_1(X;\mathbb Z) \approx \underbrace{H_1(K)}_{\mathbb Z} \oplus \underbrace{H_1(L)}_{\mathbb Z} $$ If $f(K)=L$ then the induced homology isomorphism $f_* : H_1(X;\mathbb Z)\to H_1(X;\mathbb Z)$ switches the two direct factors, so $f_*$ is not the identity, so $f$ is not homotopic to the identity.
Here is a relatively simple example: two knots that are each other's mirror images need not be ambient isotopic. Such knots are known as chiral knots. In fact, the trefoil knot is a simple example.
Clearly your homeomorphism criterion is not strong enough to distinguish these knots: taking the mirror image of $\mathbb R^3$ (or $S^3$) is a homeomorphism.