Why doesn't the $\vec{F}$ in a torque become $-\vec{F}$ when the coordinate systems are inverted? For example, for this problem:

Consider the object in Fig. 9-2. Invert the coordinate system so that $ x \rightarrow-x, y \rightarrow-y $ and $ z \rightarrow-z $. Clearly $ \overrightarrow{\mathbf{r}} \rightarrow-\overrightarrow{\mathbf{r}} $ under this transformation. What happens to $ \overrightarrow{\boldsymbol{\tau}} $ and $ \overrightarrow{\mathbf{F}} $ ?
(A) $ \overrightarrow{\boldsymbol{\tau}} \rightarrow \overrightarrow{\boldsymbol{\tau}} $ and $ \overrightarrow{\mathbf{F}} \rightarrow \overrightarrow{\mathbf{F}} $
(B) $ \overrightarrow{\boldsymbol{\tau}} \rightarrow \overrightarrow{\boldsymbol{\tau}} $ and $ \overrightarrow{\mathbf{F}} \rightarrow-\overrightarrow{\mathbf{F}} $
(C) $ \overrightarrow{\boldsymbol{\tau}} \rightarrow-\overrightarrow{\boldsymbol{\tau}} $ and $ \overrightarrow{\mathbf{F}} \rightarrow \overrightarrow{\mathbf{F}} $
(D) $ \overrightarrow{\boldsymbol{\tau}} \rightarrow-\overrightarrow{\boldsymbol{\tau}} $ and $ \overrightarrow{\mathbf{F}} \rightarrow-\overrightarrow{\mathbf{F}} $

This is the fig. 9-2:

This is a solution that I found (not sure if it's correct though):

If $ \vec{r} $ became $ -\vec{r} $, that change dosen't affect $ \vec{F} $ because the force is independent so it remains the same. But it do affect torque because they are in direct relation: $$ \vec{\tau}=\vec{r} \times \vec{F} $$ So if we put minus sign before $ \vec{r} $ that changes direction of $ \tau . $ Magnitude doesn't change, it just changes direction to the opposite of the initial torque. $$ \boxed{\vec{F} \rightarrow \vec{F} \quad \vec{\tau} \rightarrow-\vec{\tau}} $$

In the above solution, I don't understand why $\vec{F}$ doesn't become $-\vec{F}$ when the coordinate systems are inverted. I'm thinking that if we move $\vec{F}$ to the origin, it's akin to $\vec{r}$, so why is it that $\vec{r}$ needs to be changed to $-\vec{r}$ while $\vec{F}$ doesn't?

 A: You have 3 different "vectors" in this problem. $F_i$ is a vector. It doesn't have an origin, and it changes sign under coordinate inversion:
$$ P(F_i) \rightarrow -F_i $$
A point, $R_i$, doesn't live in a vector space, it lives in an affine space. The non-technical explanation is that it's a an origin, $O_i$, plus some definition of an origin, $O_i$. If you change the origin, the point doesn't change, but the vector does. To make it a vector, you look at the difference between it and the origin:
$$ r_i \equiv R_i-O_i $$
Being a vector, it changes sign under coordinate inversion:
$$ P(r_i) \rightarrow -r_i $$
The torque is not really a vector. It sometimes called a pseudo-vector and/or and axial-vector. The pseudo-vector name can mean that it depends on $O_i$...if you move that, its value changes.
The axial-vector moniker indicates that it not really vector; rather, it's the non-zero parts of an antisymmetric tensor. The tensors is formed from the available dyads:
$$ T_{ij} = r_iF_j - r_jF_i $$
That tensor only has 3 independent components, so for connivence, we write it as:
$$ \tau_i = \frac 1 2\epsilon_{ijk}T_{jk} \equiv (\vec r \times \vec F)_i $$
Being a rank-2 tensor underneath, it is even under parity:
$$ P(\tau_i) \rightarrow +\tau_i $$
A deeper dive into vectors and axial vectors leads into something called geometric algebra, and the generalization to $\mathbb{R}^N$ is handled with Clifford algebras.
A: In general, under a transformation $\mathbf{T}$ that maintains the handedness of the coordinate system you have the identity
$$ \mathbf{T} ( \boldsymbol{a} \times \boldsymbol{b} ) \equiv (\mathbf{T} \boldsymbol{a}) \times ( \mathbf{T} \boldsymbol{b} ) \tag{1}$$
The transformation of the cross product of two vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ equals to the cross product of two transformed vectors. This is true only when the transformation is orthonormal and does not change the lengths of vectors $\mathbf{T}^{-1} = \mathbf{T}^\top$.
Additionally, all vector quantities to be used in mechanics need to be expressed on the same basis vectors. So all quantities need to transform to maintain this consistency.
Additionally, the cross-product implies a certain handedness to the coordinate system. Equations involving $\times$ must also all be in the same right-hand (or left-hand) rule. But in this case, the flip transformation $\mathbf{T}$ changes the handedness of the cross product. As a result
$$ \mathbf{T} ( \boldsymbol{a} \times \boldsymbol{b} ) \equiv -(\mathbf{T} \boldsymbol{a}) \times ( \mathbf{T} \boldsymbol{b} ) \tag{2}$$
or more simply $\mathbf{T}=-1$ and $\boldsymbol{r}' = -\boldsymbol{r}$,
$\boldsymbol{F}' = - \boldsymbol{F}$ and check the cross product

*

*Preserve Handeness (1) $$ -(\boldsymbol{r}\times\boldsymbol{F}) = (-\boldsymbol{r}) \times (-\boldsymbol{F}) = \boldsymbol{r} \times \boldsymbol{F} \;\;\; \boxtimes$$

*Reverse Handeness (2) $$ -(\boldsymbol{r}\times\boldsymbol{F}) = - (-\boldsymbol{r}) \times (-\boldsymbol{F}) = -(\boldsymbol{r} \times \boldsymbol{F}) \;\;\; \checkmark$$
So under a flip transformation (2) is true, and the signs of the cross-product components are flipped.
I think maybe the op confused $-\boldsymbol{r} \times \boldsymbol{F}$ with $(-\boldsymbol{r}) \times \boldsymbol{F}$ implying that force does not flip. But there is a double negative here, with the torque being $-(-\boldsymbol{r})\times(-\boldsymbol{F})$.
A: The cross product of two ordinary vectors $\vec A$ and $\vec B$, $\vec A \times \vec B$, is not itself an ordinary vector since its direction changes when we change the handedness of the coordinate system.  An ordinary vector that has a direction independent of the coordinate system is called a polar vector.  A vector whose direction depends on the handedness of the coordinate system is called an axial vector, or a pseudovector.  Torque is an axial vector, as is angular velocity.  Position and force are polar vectors.
See a good physics mechanics textbook for further details, such as Goldstein Classical Mechanics or Symon Mechanics.
