Let $g$ be a complex orthogonal matrix,i.e.$g \in M(n,C)$ and $g^tg=I_n$.
It is known that if $\lambda$ is an eigenvalue of $g$ with corresponding eigenvector $v$ and $|\lambda|≠1$,we must have $v^tv=0$ and there must exist $w≠0$ such that $gw=\lambda^{-1}w$.
I am interested in the case satisfying that $v≠0$ is an eigenvector of $g$ with eigenvalue $1$ and $v^tv=0$. Is there another eigenvector $w≠v,w≠0$, such that $gw=w$?
As shown by the answers to your other question, there exists a complex orthogonal matrix $g$ that has at least two linearly independent eigenvectors $v$ and $w$ corresponding to the eigenvalue $1$, such that $v^tv=w^tw=0$.
Yet, there also exists a complex orthogonal matrix $g$ that has only one eigenvector $v$ (up to scaling) corresponding to the eigenvalue $1$ such that $v^tv=0$. Here is a concrete example. Let $r=\frac{1}{\sqrt{2}}$, $$ X=\pmatrix{r&0&\frac{9}{8}r\\ 0&i&-\frac12i\\ ir&0&-\frac78ir} \quad\text{and}\quad J=\pmatrix{1&1\\ &1&1\\ &&1}. $$ Then $$ X^tX =\pmatrix{r&0&ir\\ 0&i&0\\ \frac{9}{8}r&-\frac{1}{2}i&-\frac{7}{8}ir} \pmatrix{r&0&\frac{9}{8}r\\ 0&i&-\frac12i\\ ir&0&-\frac78ir} =\pmatrix{0&0&1\\ 0&-1&\frac12\\ 1&\frac12&0} $$ and $$ \begin{aligned} J^t(X^tX)J &=\pmatrix{1\\ 1&1\\ &1&1} \pmatrix{0&0&1\\ 0&-1&\frac12\\ 1&\frac12&0} \pmatrix{1&1\\ &1&1\\ &&1}\\ &=\pmatrix{1\\ 1&1\\ &1&1} \pmatrix{0&0&1\\ 0&-1&-\frac12\\ 1&\frac32&\frac12}=X^tX. \end{aligned} $$ Therefore $I=(X^t)^{-1}J^t(X^tX)JX^{-1}=(XJX^{-1})^t(XJX^{-1})$. That is, $g= XJX^{-1}$ is complex orthogonal. As the Jordan form of $g$ is $J$, its eigenspace for the eigenvalue $1$ is one-dimensional. This eigenspace is spanned by the first column of $X$, which is $v=r(1,0,i)^t$. Therefore $v^tv=0$.
If you want the numeric value of $g$, it is $$ \pmatrix{\frac{5}{4}&-ir&\frac{i}{4}\\ ir&1&-r\\ \frac{i}{4}&r&\frac{3}{4}}. $$
I have followed chapter 4 of Olga Ruff’s master thesis The Jordan Canonical Forms of orthogonal and skew-symmetric matrices: characterization and examples (whose results, albeit not entirely original, are handy) to construct this $g$.