Is there a symmetric embedding of the torus in $\mathbb{R}^3$?

Question

The torus has the following symmetry exchanging the two circle coordinates: \begin{align} s : S^1 \times S^1 &\to S^1 \times S^1 \\ (x,y) &\mapsto (y,x) \end{align} This symmetry is not really visible with the usual embedding of the torus in $\mathbb{R}^3$. So this has inspired the following question: is there an embedding $\varphi : S^1 \times S^1 \to \mathbb{R}^3$ and an homeomorphism $f : \mathbb{R}^3 \to \mathbb{R}^3$ such that $f \circ \varphi = \varphi \circ s$?

In fact, I'm quite sure that it doesn't exist, but I don't know how to prove it. Studying the homology groups of $\mathbb{R}^3 \setminus \phi(S^1 \times S^1)$ for any embedding $\phi$, I could prove that it has two path-connected components, let say $P_1$ and $P_2$. Because $S^1 \times S^1$ is compact, at least one of the two path-connected components must be bounded, let say $P_1$, and we have that $P_1 \cup \phi(S^1 \times S^1)$ is compact, while $P_2 \cup \phi(S^1 \times S^1)$ is not, so there is no homeomorphism of $\mathbb{R}^3$ exchanging the two path-connected components, so if we could prove that any homeomorphism $f$ such that $f \circ \phi = \phi \circ s$ must exchange the two path-connected components, then we could conclude that it doesn't exist. Let $x_0 \in S^1$ be any point of the circle, and $i_1 : S^1 \hookrightarrow S^1 \times \{x_0\}$ and $i_2 : S^1 \hookrightarrow \{x_0\} \times S^1$ the two obvious embeddings of the circle in the torus $S^1 \times S^1$. I think that the map $\phi \circ i_1$ must be non-trivial in either $P_1 \cup \phi(S^1 \times S^1)$ or $P_2 \cup \phi(S^1 \times S^1)$, but not in both, and $\phi \circ i_2$ must be non-trivial in the other one. If this is true, we could conclude, but I don't know how to prove this.

Answer 1

If the embedding is locally the standard embedding of the $2$-disk in the $3$-disk then we can define orientations using local homology and follow ideas from the comments to the question. Since $s$ switches the orientation of the torus $T = S^1 \times S^1$, $f$ must switch the two components of $\mathbb{R}^3 \setminus \varphi(T)$. Since $\varphi(T)$ is compact and hence bounded, exactly one of these two components is bounded. But a subset of $\mathbb{R}^3$ is bounded iff its closure is compact, so this is a property preserved by homeomorphisms.