Let $D_n=\{x,...,x^n,y,yx,...,yx^n\}$. I have already proven that we can distinguish some isomorphism classes (of two dimensional representations) by saying they map $x$ to matrices of the form $\begin{pmatrix} e^{2\pi ik/n}&&0\\0&&e^{-2\pi ik/n}\\ \end{pmatrix}$ where $k\in\{1,...,n/2-1\}$ if $n$ is even and $k\in\{1,...,(n-1)/2\}$ if $n$ is odd. I have also proven that $y$ must map to either $\begin{pmatrix} 0&&1\\1&&0\\ \end{pmatrix}$ or $\begin{pmatrix} 0&&-1\\-1&&0\\ \end{pmatrix}$.
According to this https://groupprops.subwiki.org/wiki/Linear_representation_theory_of_dihedral_groups#The_two_one-dimensional_representations, the classes which map $y$ to the negative matrix are isomorphic to something, but I don't see how that's possible.
My textbook defines an isomorphism as a transformation $T$ between the two vector spaces $V,V'$ of each representation such that for any $v\in V$ and any $g$ in our group ($D_n$), $gT(v)=T(gv)$. Using that definition, for a representation with positive $y$ matrix to be isomorphic to a representation with negative $y$ matrix, we would need $yT(v)=T(-yv)$ where now the two $y$ represent that same matrix. But this is impossible, right? Am I missing something?