Can one disjoin any submanifold in $\mathbb R^n$ from itself by a $C^{\infty}$-small isotopy?

Here's a counter example. Take any embedding of $\mathbb{C} P^2$ in $\mathbb{R}^7$ (such embeddings exist by

Steer, B., On the embedding of projective spaces in Euclidean space, Proc. Lond. Math. Soc., III. Ser. 21, 489-501 (1970). ZBL0206.25501;

the construction is summarised in this answer on MSE).

If such an embedding admits a normal vector field, then by the Compression Theorem of Rourke and Sanderson it is isotopic to an embedding with normal field parallel to the last coordinate of $\mathbb{R}^7=\mathbb{R}^6\times\mathbb{R}$, and then the projection is an immersion $\mathbb{C} P^2\looparrowright \mathbb{R}^6$. Such immersions cannot exist, by András Szűcs' answer here.

There are lots of counter-examples.

Here's one: The fibration $\text{SO}(3) \to \text{SO}(4) \to S^3$ splits, since it is the principal bundle of the tangent bundle of $S^3$, and the latter is parallelizable. In particular, $\pi_5(\text{SO}(4)) \cong \pi_5(\text{SO}(3)) \oplus \pi_5(S^3) \cong \Bbb Z/2 \oplus \Bbb Z/2$ .

Let $a$ and $b$ denote the generators for the summands. Then $(a,b)$ determines a rank $4$-vector bundle over $S^6$ with trivial Euler class (since $H^4(S^6) = 0$). But this vector bundle does not have a section, for if it did that would lead to the contradiction that $b=0$.