Identity $\mathbb{C}^n$ with $\mathbb{R}^{2n}$ via the equality $$(z_1, z_2, \ldots, z_n)=(x_1, \ldots, x_n, y_1, \ldots , y_n)$$
Where $z_j=x_j + iy_j$.

We call a linear invertible map $A: \mathbb{R}^{2n}\to \mathbb{R}^{2n}$ totally real if the image of every complex line is not a complex line (i.e a totally real plane). Equivalently, if $Gr(2,2n)$ denotes the Grassmannian of planes in $\mathbb{R}^{2n}$ then  $\mathbb{C}P(n-1)$ lives inside $Gr(2,2n)$, and $A$ being totally real means that the image of $\mathbb{C}P(n-1)$ intersects itself trivially. Note that the (real) dimension of $\mathbb{C}P(n-1)$ is half the dimension of $Gr(2,2n)$.

As an example, for $n=2$, the linear map $$B:(x_1, x_2, y_1, y_2)\to (-y_2
, x_1, y_1, -x_2)$$
Is totally real and satisfies $\det(B)=1$. Since $B$ is isotopic to identity through linear maps, then $B(\mathbb{C}P(1))$ is isotopic to $\mathbb{C}P(1)$ which shows that the self intersection number of $\mathbb{C}P(1)$ (inside $Gr(2,4)$) is zero.

Similarly, the previous example can be generalized to all $n=2k$ even.

My question: is it possible to find a totally real linear (invertible) transformation $A$ for $n\geq 3$ odd? (we allow $A$ to have negative determinant).

Note that if the self intersection number of $\mathbb{C}P(n-1)$ (inside $Gr(2,2n)$) is non-zero for $n$ odd, then $A$ cannot have positive determinant. But I don't know if this is the case and I would like also to know what is the self intersection number in this case.

Thanks in advance.