Is there a "non-trivial" subgroup $G$ of $GL(n,\mathbb C)$ whose every element has an eigenvalue $1$?
"Non-trivial" here means that is no common eigenvector with eigenvalue $1$ for all elements of $G$.
Edit: As pointed out by Noam D. Elkies, $SO(n,\mathbb C)\subset GL(n,\mathbb C)$ has this property. The above was a simplified (and clearly oversimplified) version of my actual problem: If $G\subset SO(2n,\mathbb C)$ has the property that its every element has an eigenvalue $\pm 1$ then is there a common eigenvector with eigenvalue $\pm 1$ for elements of $G$? [I think it is true for every embedding of $G=SO(2n-1,\mathbb C)$ into $SO(2n,\mathbb C)$.]
Edit: As Elkies says, the symmetric power of the defining rep of $SO(5,\mathbb C)$ gives an irred rep into $SO(14,\mathbb C)$, and all elements of its image have $1$ as its eigenvalue.
Hence perhaps one can ask: Is there an irreducible subgroup of $SO(2n,\mathbb C)$ for $n<7$ whose all elements have eigenvalue $1$? ("Irreducible" in the sense of being a representation on $\mathbb C^{2n}$.)
