There are various different notions of knot, and of equivalence of knots. For polygonal knots, there is combinatorial equivalence (extend an edge out into two sides of a triangle), equivalence via orientation preserving homeomorphism, and equivalence via ambient isotopy. These notions of equivalence are all equivalent, shown by a result of Moise [1]. These notions extend immediately to continuous knots that are equivalent to polygonal knots under ambient isotopy or orientation preserving homeomorphism.
Then for smooth knots, the obvious notions of equivalence are equivalence by smooth isotopy, equivalence by smooth ambient isotopy, and equivalence by orientation preserving diffeomorphism. These notions of equivalence are again equivalent, by an isotopy extension theorem ([2], Chapter 8, theorem 1.3) and a result of Cerf [3].
One can extend the former kind of equivalence to smooth knots - if $S$ and $S'$ are smooth knots, then there are polygonal knots $J$ and $J'$ with $S$ ambiently isotopic to $J$, and $S'$ ambiently isotopic to $J'$. The question of whether $J$ is equivalent to $J'$ in the first sense then gives us a notion of equivalence for smooth knots. Call this "polygonal equivalence" for smooth knots. Call the other kind of equivalence "smooth equivalence".
Obviously if $S$ and $S$ are smoothly equivalent, then they are polygonally equivalent (by composing ambient isotopies). But is there a good reference for the converse?
Manturov [4] cites Crowell and Fox [5] in this connection, in section 2.1, but they do not appear to provide a proof. The great majority of knot theory textbooks seem to just take the polygonal approach.
There may be general theorems about homeomorphisms being isotopic to diffeomorphisms relative subspaces that would do the trick, taking you from an orientation preserving homeomorphsim $F:\mathbb{R}^3\to\mathbb{R}$ with $F(S)=S'$ to an orientation preserving diffeomorpism $F:\mathbb{R}^3\to\mathbb{R}^3$ with $F(S)=S'$. I do not know of such results however.
Otherwise I believe one could give a direct proof, since if $S$ and $S'$ are polygonally equivalent then there is a piecewise smooth isotopy from the first to the second, and I believe this can be smoothed out. But presumably this has already been done by someone (or the result deduced from more general theorems).
[1] Moise, Edwin E., Affine structures in 3-manifolds. VIII. Invariance of the knottypes; local tame imbedding, Ann. Math. (2) 59, 159-170 (1954). ZBL0055.16804.
[2] Hirsch, Morris W., Differential topology, Graduate Texts in Mathematics. 33. New York - Heidelberg - Berlin: Springer-Verlag. X, 221 p. DM 36.20; $ 14.80 (1976). ZBL0356.57001.
[3] Cerf, J., Sur les difféomorphismes de la sphère de dimension trois $\Gamma\sb 4 = 0$, Lecture Notes in Mathematics. 53. Berlin-Heidelberg-New York: Springer-Verlag. XII, 133 p. (1968). ZBL0164.24502.
[4] Manturov, Vassily, Knot theory, Boca Raton, FL: Chapman & Hall/CRC (ISBN 1-415-31001-6/hbk). 400 p. (2004). ZBL1052.57001.
[5] Crowell, R.H.; Fox, R.H., Introduction to knot theory, Moskau: Verlag ’Mir’. 348 S. (1967). ZBL0149.20901.