(This question is my assumption, I'm not sure if it's true.)
I have already proven cases where $n$ is odd. Since the characteristic polynomial $p_T(t)$ of $T$ on $R^n$ is in the form $$p_T(t) = t^n + \dots + \det([T]_\beta)$$ Since $n$ is odd, we have the sign of $\lim_{t\to -\infty}p_T(t)$ is different from $\lim_{t\to \infty}p_T(t)$, which implies that $p_T(t)$ has a solution on $R$. Thus, $T$ has an eigenvalue, which means there exists an eigenvector.
However, I have no idea about cases when $n$ is even.
When $n=2$, the rotation matrix is a counterexample.
When $n>2$, intuitively the orthogonal operator on $R^n$ is like rotation on a 2-dimension plane, which kinda implies that exists vector $v$ in the remaining $n-2$ dimensions, which is not affected by that rotation (Then $Tv = \pm v$).
But I can't find any relationship between orthogonal and even dimensions. Can anyone give me a hint?
FYI: I comes up with this question when I'm proving
"Suppose $S\in \mathcal{L}(R^3)$ is orthogonal; Prove that $\exists x\in R^3, x\neq 0, s.t., S^2 x = x$",
which is an exercise in linear algebra done right.
With the help of @totoro, the conjecture has been disproved.
Consider $\begin{pmatrix}\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0&0\\\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0&0\\ 0&0&\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\0&0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{pmatrix}$.