TITLE: Which two knots are isotopic but not ambient isotopic?
QUESTION [6 upvotes]: Which two knots are isotopic but not ambient isotopic? How can we see that they are indeed not isotopic but not ambient isotopic?

REPLY [27 votes]: The following animation shows an isotopy from a knotted curve to a straight segment. If we were to connect the two ends together, this would give an isotopy from a trefoil knot to an unknot.  The same trick works with any tame knot, which means that basically any two knots are isotopic.

The knotted portion of the curve shrinks linearly at $t$ goes from $0$ to $1$. Note that this really is an isotopy -- the curve doesn't intersect itself for any $t<1$, and it doesn't intersect itself at $t=1$ either.
What this means is that isotopy of embeddings really doesn't really capture what it means for two knots to be the same, so we need use some other criterion to define equivalence of knots.  There are several standard solutions to this problem:

Use ambient isotopy, as mentioned by the OP.  Although the animation above shows an isotopy of embeddings of a circle into $\mathbb{R}^3$, this doesn't extend to an isotopy of homeomorphisms $\mathbb{R}^3\to\mathbb{R}^3$.  The problem is that some of the substance of $\mathbb{R}^3$ gets "squeezed together" when the knot shrinks to a point.
Use smooth isotopy. The isotopy above isn't differentiable at the point where the knot disappears.
Use thick tubes instead of circles.  This makes it impossible to tighten a knot to a single point.