If $u$ is a linear orientation preserving isometry on an inner product sapce of odd dimension, then $1$ is an eigen value of $u$.
Let $D$ be the diagonal matrix $diag\{\e_1,\ldots,\e_n\}$, $\e_j=\pm 1$, and $A$ the matrix of $u$ with respect to an orthonormal basis $e_1,\ldots,e_n$ satisfying $\la e_j,e_j\ra=\e_j$. Since $u$ is an isometry, we have $u^*=u^{-1}$ and thus $DA^tD=A^{-1}$ and since $u$ is orientation preserving: $\det A=1$. Finally it follows from $D^2=1$ that
\begin{eqnarray*}
\det(A-1)
&=&\det(1-A^{-1})
=\det(1-DA^tD)\\
&=&\det(D(1-A^t)D)
=\det(1-A^t)
=\det(1-A)
=-\det(A-1)~.
\end{eqnarray*}