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?



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


