The question is:
Let $T(x, y, z) = (x, y , z)$ and $K_{1}$,$K_{2}$ are 2 knots such that $T(K_{1}) = K_{2}$. Show that $K_{1} \cong K_{2}$.
Where $K_{1} \cong K_{2}$ is defined below:
The definition of equivalence of 2 knots according to Richard H. Crowell and Ralph H. Fox, edition 1963, is:
Assume that $K_{1}$,$K_{2}$ are 2 knots in $\mathbb{R^3}$, then they are equivalent , denoted by $K_{1} \cong K_{2}$, iff $\exists f: \mathbb{R^3} \rightarrow \mathbb{R^3}$, where $f$ is a homeomorphism and such that $f(K_{1}) = K_{2}.$
The hint that was given to me is:
The solution is an easy consequence of this question:
Showing that two knots are equivalent if an invertible linear transformation maps one onto another.
But I do not understand how, could anyone clarify this for me please?
Suppose that we know that the linear map $f \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$ is bounded. Then we get $$\left\|f(v) - f(v')\right\|= \left\|f(v - v')\right\| \leq C\left\|v-v'\right\|$$ for some $C$. Thus $f$ is Lipschitz continuous and therefore also continuous. Therefore it suffices to show that any linear map $f \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$ is bounded. Let $v \in \mathbb{R}^n$ and write $v = v_1e_1 + \dots + v_n e_n$ as a linear combination of the standard basis. Then we get $$\left\|f(v)\right\| = \left\|v_1f(e_1) + \dots + v_n f(e_n)\right\| \leq \vert v_1 \vert\left\|f(e_1)\right\| + \dots + \vert v_n \vert \left\| f(e_n)\right\|$$ by the triangle inequality. As any two norms on $\mathbb{R}^n$ are equivalent, there exists some $C$ such that $\vert v_1 \vert + \dots + \vert v_n \vert \leq C \left\| v\right\|$. Thus we get $$\left\| f(v)\right\| \leq \vert v_1 \vert\left\|f(e_1)\right\| + \dots + \vert v_n \vert \left\| f(e_n)\right\|\leq (\vert v_1 \vert + \dots + \vert v_n \vert)D \leq CD \left\| v\right\|,$$ where $D = \text{sup}(\left\|f(e_i)\right\|)$. Thus $f$ is bounded.
This means you can take the invertible linear map as the homeomorphism that you were searching.