Does it have a physical interpretation whether the determinant of a Lorentz transformation is $1$ or $-1$? It is known that Lorentz transformations have either the determinant $1$ or $-1$ (one of the answers to this question gives a proof). Does the determinant have a physical meaning?
Motivation (for those who are familiar with the mathematical formulation of SR):
Let $M$ be the Minkowski space and let $x\colon M\to\mathbb R^4$ be a chart with the property
\begin{equation*}
    g_{ij}|_p=g_p\left(\frac{\partial}{\partial x^i}|_p,\frac{\partial}{\partial x^j}|_p\right)=\begin{pmatrix}
-1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1
\end{pmatrix}
\end{equation*}
It can be proven that another chart $y\colon M\to \mathbb R^4$ has the same property if and only if $x\circ y^{-1}$ is a Poincarré-Transformation. If $y$ is such a chart, then the basis change matrix for the tangent space is a Lorentz transformation in each point. That is, $x$ and $y$ determine the same orientation if and only if that Lorentz transformation has determinant $1$. Thus, a probably equivalent question would be whether orientation has a physical meaning in this case.
 A: Mathematically speaking, you're here talking about the representation theory of the Lorentz group O(3,1). This group has four connected components, depending on the sign of the 00 element and the sign of the determinant.
The ones with the positive determinant are called proper Lorentz transformations. The ones with positive determinant and positive 00 component are called the proper orthochronous subgroup.
Note that the improper transformations alone do not form a subgroup (because det($\Lambda_1\Lambda_2$) = det($\Lambda_1$)det($\Lambda_2$)), and the same is true for non-orthochronous transformations. So if you include Lorentz transformation with negative determinant, you are effectively only including one additional generator in your Lorentz group, the one that reverses the sign of the determinant. You still need to have all the proper orthochronous transformations present.
So what's the physical interpretation of negative determinants? Well, it depends on which generator you choose to generate the negative determinant.
One option is to include the operator $T$ which reverses time, which physically corresponds to adding a symmetry $t\rightarrow -t$ to your system (in addition to Lorentz invariance). This takes you to the improper non-orthochronous component.
Another option is to include the operator $P$ which reverses spatial coordinates, e.g. $\mathbf{x}\rightarrow -\mathbf{x}$. This takes you to the improper orthochronous component.
There is a final option, which is to include the combined symmetry $PT$, which takes you to the proper non-orthochronous component, but this has positive determinant.
So, if your symmetry group includes Lorentz transformations with negative determinant, then they either correspond to a transformation taking $t\rightarrow -t$ or $\mathbf{x}\rightarrow -\mathbf{x}$ (possibly combined with a proper orthochronous transformation).
Note that only the proper orthochronous component is smoothly connected to (and includes) the identity, so in order to do write e.g. infinitesimal boosts, one usually works exclusively with the proper orthochronous component of the group (at least in e.g. quantum field theory).

As for your final point about orientability: This is related to the topological study of spacetime. In particular, there's a topological invariant (the first Stiefel-Whitney class), which characterizes exactly whether or not your system is orientable. The Stiefel-Whitney class actually also distinguishes different phases of matter in non-relativistic quantum physics. This is an example of a toplogical phase. Mathematically, this corresponds to the difference between O(N) symmetry and SO(N) symmetry.
