A reflected sphere embedding is regularly homotopic but not isotopic to the standard one

Both $\iota$ and $r\circ \iota$ are smooth embeddings of $S^2$ into ${\mathbb{R}}^3$: $\iota$ is the inclusion of an embedded submanifold, and $r$ is a linear isomorphism, hence a diffeomorphism of ${\mathbb{R}}^3$ whose composite with $\iota$ is again an embedding.

The two embeddings are regularly homotopic: by [L1], applied with $m=2$ and $n=3$, all immersions $S^2\to {\mathbb{R}}^3$ are regularly homotopic because ${\pi }_2(SO(3))=0$, and $\iota$ and $r\circ \iota$ are such immersions; the reflection is, up to an orientation-preserving rotation of the target, the antipodal reparametrisation of the standard inclusion, and the direct formal-data computation identifies the obstruction as a class in ${\pi }_2(SO(3))=0$, so the two formal data are homotopic and the formal-data criterion gives the regular homotopy. Hence clause 1 of the counterexample holds.

The two embeddings are not isotopic. Suppose an isotopy of embeddings from $\iota$ to $r\circ \iota$ existed. Since $S^2$ is compact and ${\mathbb{R}}^3$ is boundaryless, [L2] produces an ambient isotopy $H$ of ${\mathbb{R}}^3$ with $H_0=\mathrm{i}\mathrm{d}$ and $H_1\circ \iota =r\circ \iota$, that is $H_1{\mathrm{\mid }}_{S^2}=r{\mathrm{\mid }}_{S^2}$.

The sphere complement has precisely the two connected components $U=\{\mathrm{\mid }x\mathrm{\mid }<1\}$ and $V=\{\mathrm{\mid }x\mathrm{\mid }>1\}$: $U$ is convex, and in $V$ radial paths to a common large sphere followed by great-circle arcs on that sphere (for antipodal endpoints choose a perpendicular unit vector by normalizing the first nonzero coordinate-vector projection) connect any two points. Since $H_1(S^2)=S^2$, the homeomorphism permutes these components. It cannot send $U$ to $V$, since $H_1(\bar{U})$ is compact and therefore bounded, whereas $V$ is unbounded. Consequently $H_1(U)=U$ and $H_1(B^3)=B^3$.

For every $x$, the function $t\mapsto \det d{H}_t(x)$ is continuous, never zero, and equals one at zero, so [L4] makes it positive for all $t$. Thus $H_1$ preserves the ambient orientation. Because it maps the ball's interior onto itself, its differential takes an outward transverse vector to an outward transverse vector at the sphere: in a boundary chart the inward normal coordinate has positive inward derivative, by invertibility and preservation of the interior. The outward-normal-first rule in [L4] therefore makes $H_1{\mathrm{\mid }}_{S^2}$ orientation preserving. By [L3] its degree is $+1$, contradicting the reflection's degree $-1$. This proves the orientation argument locally, without a B-page prerequisite.

Therefore $\iota$ and $r\circ \iota$ are regularly homotopic but not isotopic, so the statement refuted is false: regular homotopy of embeddings is strictly coarser than isotopy of embeddings for $S^2$ in ${\mathbb{R}}^3$. The historically first instance, a knotted circle versus the round circle in ${\mathbb{R}}^3$, is recorded as a boundary rather than proved here, because its non-isotopy invariant ${\pi }_1({\mathbb{R}}^3\setminus L)$ belongs to the low-dimensional knot track and not to this run's closure.