Question 8.6.6 from Groups, Matrices and Vector Spaces by James Carrell:
The 4-cube has eight diagonals. They are represented by the semidiagonals $\pm \mathbf{e}_1 + \pm \mathbf{e}_2 + \pm \mathbf{e}_3 + \mathbf{e}_4$. Find a pair of diagonals $D_1$ and $D_2$ such that no $\sigma \in \textrm{Sym}(C(4))$ satisfies $\sigma(D_1) = D_2$.
I'm confused since given any pair of semidiagonals, we can construct an orthogonal map from one corner to another. Then this linear map also works for the opposite sides of the diagonal too.
Let $\mathbf{v}_1$ be the semidiagonal for the diagonal $D_1 = \{ u \mathbf{v}_1 | -1 \leq u \leq 1 \}$ and likewise for $D_2$. We construct a linear map $T(\mathbf{v}_1) = \mathbf{v}_2$ which is orthogonal. Then also $T(-\mathbf{v}_1) = -\mathbf{v}_2 \implies T(D_1) = D_2$.
What's wrong with my reasoning and how do I answer the question?