Does there exist a closed smooth manifold $M$ and a diffeomorphism $f\colon M \to M$ such that:
If the answer is yes, what is an example?
Such examples exist in dimension 5, they are contained in the paper by Cameron Gordon "On the higher-dimensional Smith conjecture", Proc. London Math. Soc. (3) 29 (1974), 98–110. Namely, Gordon proves in Theorem 5 of this paper that (for every $n\ge 5$) there are infinitely many smooth knots $K=S^{n-2}\subset S^n$ so that $K$ is the fixed-point set of a ${\mathbb Z}_p$-action $\alpha_p$ for every prime $p$. He also proves (Theorem 4) that, given $K$, if every action $\alpha_p$ extends to a circle action on $S^n$ then $\pi_1(S^n\setminus K)\cong {\mathbb Z}$. He then notes (a theorem by Levine) that, for $n\ge 5$, if $\pi_1(S^n\setminus K)\cong {\mathbb Z}$ then $K$ is smoothly unknotted in $S^n$. Since Gordon's theorem 5 yields infinitely many smooth isotopy classes of knots $K$, it then follows for every such (nontrivial) knot, at least for some prime $p$, one of Gordon's actions $\alpha_p$ does not extend to a smooth circle action.
Finally, every diffeomorphism of $S^5$ is PL isotopic to the identity (by the Alexander's trick). Since in dimensions $<7$, PL=DIFF, we conclude that the generator of $\alpha_p({\mathbb Z}_p)$ is smoothly isotopic to the identity.
Lastly, note that in the topological category, the examples exists already for $M=S^3$: Bing ("Inequivalent Families of Periodic Homeomorphisms of $E^3$", Ann. of Math. 80 (1964) 78-93.) constructed finite order homeomorphisms whose fixed-point sets are wild knots, while F.Raymond (Classification of the actions of the circle on $3$-manifolds. Trans. Amer. Math. Soc. 131 (1968) 51-78.) proved that the every $S^1$-action on $S^3$ is topologically conjugate to an orthogonal action.
