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.