This should be true, I think. Since $T$ maps $E\cong \mathbb{R}^{n-1}$ into itself, what must be shown is the following:
If $T\colon \mathbb{R}^{n-1}\to \mathbb{R}^{n-1}$ is an orientation preserving orthogonal linear transformation ($n$ even), then there is a $v\in \mathbb{R}^{n-1}$ such that $Tv = v$.
Let's prove this by contradiction. If it were false, then the linear map $T - \mathrm{Id}$ would be invertible. I claim that, in fact, for every $\lambda\in [0,1]$, the linear map $\lambda T - (1-\lambda)\mathrm{Id}$ is invertible. The only way this could be false is if there is a $v\in \mathbb{R}^{n-1}$ such that $\lambda Tv = (1-\lambda)v$, or in other words, that $Tv = \frac{1-\lambda}{\lambda} v$. Since eigenvalues of orthogonal transformations have absolute value $1$, this can only possibly happen when $\lambda = 1/2$. But $(1/2)T - (1/2)\mathrm{Id} = (1/2)(T - \mathrm{Id})$ is invertible since $T - \mathrm{Id}$ is invertible. This proves $\lambda T - (1-\lambda)\mathrm{Id}$ is invertible for all $\lambda\in [0,1]$. Let $\varphi(\lambda) = \det(\lambda T - (1-\lambda)\mathrm{Id})$. Then $\varphi$ is a continuous nonvanishing function on $[0,1]$. But $\varphi(1) = \det T = 1$ and $\varphi(0) = \det(-\mathrm{Id}) = (-1)^{n-1} = -1$, so $\varphi$ changes sign on $[0,1]$, a contradiction of the fact that $\varphi$ is nonvanishing.