# Fixed points of diffeomorphisms of tori isotopic to identity and their traces under isotopies

Suppose $$T^n$$ is the $$n$$-dimensional torus ($$n\geq 2$$) and $$f: T^n\to T^n$$ is a diffeomorphism isotopic to the identity and fixing points $$x_1,\ldots,x_k\in T^n$$. Does there exist an isotopy $$\{ f_t: T^n\to T^n\}_{0\leq t\leq 1}$$ connecting $$f_0=Id$$ with $$f_1=f$$ so that all the loops $$\{ f_t (x_i)\}_{0\leq t\leq 1}$$, $$i=1,\ldots,k$$, lie in the same free homotopy class?

I believe this is not always possible: let $$d\colon \mathbb R \to [0,1/2]$$ send a real number to the distance to the nearest integer. Consider the map $$F\colon \mathbb R^2, (x,y) \mapsto (x+2d(y),y),$$ which commutes with the $$\mathbb Z^2$$ action on $$\mathbb R^2$$ and thus descends to a homeomorphism of $$T^2 = \mathbb Z^2 \backslash \mathbb R^2$$, denoted $$f$$. Of course, $$f$$ is not smooth, but we can clearly make it smooth it by slightly changing the formula above (replace $$d$$ by a function that is smooth and takes values $$0$$ on integers and $$1/2$$ on half-integers).
Geometrically, $$f$$ can be understood as a Dehn twist on half of the torus and the reverse Dehn twist on the other half.
Clearly, $$f$$ perseveres $$p= [(0,0)]$$ and $$q = [(0,1/2)]$$ and is isotopic to the identity.
We can lift any isotopy $$f_t$$ with $$f_0 = Id$$ and $$f_1 = f$$ to an isotopy $$F_t$$ of $$\mathbb R^2$$ from $$F_0$$, a translation by, say, $$(z_1,z_2) \in \mathbb Z^2$$, to $$F_1 = F$$ from above. But then $$F_t(p)$$ goes from $$(z_1,z_2)$$ to $$(0,0)$$, whereas $$F_t(q)$$ goes from $$(z_1,z_2+1/2)$$ to $$(1,1/2)$$, so the loops these curves describe in $$T^2$$ are not freely homotopic.