of a projection of $f(M_{2N+1})$ will change the signed crossing number between the two edges by $\pm 2$. Thus any crossing change will alter $\widehat{L}_{\varepsilon}(f)$ by an even number. Now consider the embedding $f$ of $M_{2N+1}$ shown in Figure \[FigureM2N1\_mobius\_embedding\_labeled\]. There is only one crossing, and it is between two outer edges with an outer edge distance of $2N$ (the maximum). The epsilon coefficient for this crossing is $-1$, which is multiplied by the crossing sign $-1$ so that $\widehat{L}_{\varepsilon}(f) = (-1)(-1) = 1$ for this embedding. It follows that $\widehat{L}_{\varepsilon}(f)$ is odd for any embedding of $M_{2N+1}$.

![An embedding of $M_{2N+1}$ with $\widehat{L}_{\varepsilon}(f) = 1$.[]{data-label="FigureM2N1_mobius_embedding_labeled"}](M2N1_mobius_embedding_labeled){width=".45\textwidth"}

\[mobius automorphism\] Let $N \geq 2$. If $\alpha$ is an automorphism of $M_{2N+1}$, then the epsilon coefficients of $M_{2N+1}$ and $\alpha(M_{2N+1})$ are the same, and $\alpha$ either preserves the orientation of every edge or reverses the orientation of every edge.

Let $\alpha \in$ Aut$(M_{2N+1})$. Since $N \geq 2$, it follows from Simon [@S86] that $\alpha$ takes the cycle $\overline{x_1 x_2 ... x_{4N+2}}$ to itself, preserving the order of the edges $x_1, x_2, ..., x_{4N+2}$ and thus preserving the order of the edges $y_1, y_2, ..., y_{2N+1}$ as well. Because the order of the outer 