Question: Is every $ \mathbb{R}P^{2n} $ bundle over the circle trivial?

Are there exactly two $ \mathbb{R}P^{2n+1} $ bundles over the circle?

This is a cross-post of (part of) my MSE question

<https://math.stackexchange.com/questions/4349052/diffeomorphisms-of-spheres-and-real-projective-spaces>

which has been up for a couple weeks and got 8 upvotes and some nice comments but no answers.

My intuition for thinking both answer are yes is that there are exactly 2 sphere bundles over the circle. The trivial one and then the non-trivial (and non-orientable) one which can be realized as the mapping torus of an orientation reversing map of the sphere. So importing that intuition to projective spaces then the orientable $ \mathbb{R}P^{2n+1} $ should have a nontrivial (and non orientable) bundle over the circle while the non orientable $ \mathbb{R}P^{2n} $ should have only the trivial bundle. For $ n=1 $ this checks out since that projective space is orientable and thus we have exactly two bundles over the circle (the trivial one=the 2 torus and the nontrivial one=the Klein bottle).

Answer: 
It is not possible with 7 (i.e., with a triangle $T$ and a quadrilateral $Q$). I write a rough proof.

First, any quadrilateral $Q$ lying in a plane $\pi$ can be partitioned in two triangles $Q\_1$ and $Q\_2$, whose common edge is a diagonal $d$ of $Q$. Now the intersection of the triangle $T$ with $\pi$ consists of two points (otherwise they are coplanar and the conclusion is trivial). Since the polygons are not linked, there are two cases: either both points lie outside $Q$, or they both lie inside $Q$.

Case 1: both lie outside. An inspection of the cases shows that we can just translate $T$ in direction parallel to one of the bisectors of the triangles $Q\_1$ or $Q\_2$ starting from one of the vertices not belonging to $d$.

Case 2: both lie inside. We will consider a modified problem: given a quadrilateral $Q$ and two initial points $x(0),y(0)$ inside it, find two continuous curves $x(t),y(t)$ such that the distance between the two is decreasing in time to 0, and they never lie in $Q$. The solution is, e.g., a linear homotopy that sends both points to the midpoint of $d$. Now we only have to realize $x(t)$ and $y(t)$ as the intersections of $T$ and the plane $\pi$. One can convince themselves that this is possible by pulling $T$ in direction orthogonal to $\pi$ while translating it.
I think that 8 might be possible, by interlocking two Star Trek symbols as shown below.

[![enter image description here](https://i.stack.imgur.com/RyurZ.png)](https://i.stack.imgur.com/RyurZ.png)

**Adendum:** This candidate may not work, as quarague points out, but I leave it as a potential "how not to" example. There are other ways to cross the two figures, while they remain unlinked, and one of these variations could be more promising. Also I think it is important to specify whether the loops have any thickness or not. Some configurations may interlock assuming there is nonzero thickness.