$\newcommand{\R}{\mathbf R}$ The question below is motivated by the fact that each non-identity member of $SO(3)$ fixes exactly one line pointwise in $\R^3$.
Let $f:\R\to SO(3)$ be a smooth map (which models a smoothly rotating pivoted rigid body in $\R^3$).
Question. Does there necessarily exist a smooth map $g:\R\to \mathbf RP^2$ such that $f(t)$ fixes $g(t)$ pointwise for each $t$.
The question above has negative answer if we do not require $f$ to be smooth but only continuous. For then we can take $f$ to be as follows: for $t\in [0, 1]$ we define $f(t)$ as rotation about the $z$-axis by $2\pi t$, and for $t\in [1, 2]$ we deifne $f(t)$ as rotation about the $x$-axis by $2\pi (t-1)$. For all other $t$ we define $f(t)=I$. Now if there exists a continuous $g:\R\to \R P^2$ such that $f(t)$ fixes $g(t)$, then for $t\in (0, 1)$ we must have $g(t)$ is the $z$-axis, and for $t\in (1, 2)$, $g(t)$ equals the $x$-axis. Thus $g(1)$ cannot take any value without destroying continuity.
Of course, this counterexample does not work for smooth case.