Which two knots are isotopic but not ambient isotopic? Which two knots are isotopic but not ambient isotopic? How can we see that they are indeed not isotopic but not ambient isotopic?
 A: In the topological category, all embeddings of tame knots (and, I believe, wild knots with a tame arc) are isotopic to the unknot (and hence to one another by symmetry).  The idea is to shrink everything outside of a nice arc down to a point, so you get the original knot (with a smaller and smaller "knotted" part) for all time $t \in [0,1)$ and then the unknot at $t=1$.
When you add in a standard assumption like locally-flat or smooth embeddings (and isotopies, where locally-flat means the embedding $S^1 \times I \hookrightarrow S^3 \times I$ arising from the isotopy is locally flat), then the isotopy extension theorem says that all isotopies of the embedding extend to ambient isotopies. See this post and the answer by Mike Miller I referenced above.
So we have lots of examples of knots which are isotopic but not ambient isotopic: any two knots that are not ambient isotopic (e.g. the unknot and the trefoil) but each contain at least one tame arc will be isotopic.
A: 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.
