The round circle and its reflection are not isotopic embeddings in the plane

Both maps are embeddings. The radial fields $n_i(x)=x$ and $n_r(x)=r(x)$ are smooth nowhere-zero normal vectors: reflection preserves the inner product, so $⟨r(x),d{r}_x(v)⟩=⟨x,v⟩=0$ for tangent vectors $v$. They trivialize the normal lines. Both double point sets are empty, so self-transversality is vacuous and [F3] gives zero primary counts.

Suppose an isotopy exists. By [F1] its ambient extension has time-one diffeomorphism $H$ with $H\circ i=r$. Its orientation sign is positive, since the differential determinants of the ambient isotopy vary continuously from the identity and never vanish. Also $H(S^1)=S^1$. The plane complement consists of the open disk and its exterior, and $H$ permutes these components. The image of the closed disk is compact; thus the open disk cannot map to the exterior, whose closure is unbounded. Therefore $H(D^2)=D^2$.

The orientation-preserving disk diffeomorphism $H{\mathrm{\mid }}_{D^2}$ carries outward-pointing boundary vectors to outward-pointing vectors: in local boundary coordinates a diffeomorphism preserving the interior has positive inward-coordinate derivative on the boundary. Consequently it preserves the induced boundary orientation, so $H{\mathrm{\mid }}_{S^1}$ has degree $1$ by [F2]. But $H\circ i=r$ identifies this restriction with the coordinate reflection of degree $-1$, a contradiction. Thus the embeddings are not isotopic, while their normals and primary counts agree by step 1.1.