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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 $\mathbb C^{2n}$.)

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    $\begingroup$ Hmm, I did not see unipotent mentioned. Couldn't one take the complement to the trivial rep. in any linear rep. arising in the natural manner from a transitive subgroup of $S_n$ containing no $n$-cycle? $\endgroup$
    – John Shareshian
    Commented Mar 11, 2015 at 2:13
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    $\begingroup$ ${\rm SO}_n$, for $n>1$ odd? $\endgroup$
    – Noam D. Elkies
    Commented Mar 11, 2015 at 2:13
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    $\begingroup$ Every connected compact simple Lie group acts on its adjoint representation with the property that $1$ is an eigenvalue for every element. $\endgroup$
    – Venkataramana
    Commented Mar 11, 2015 at 2:16
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    $\begingroup$ There are still counterexamples in ${\rm SO}_{2n}$. For example, a symmetric power of ${\rm SO}_m$ for $m$ odd (e.g. if $m \equiv 1 \bmod 4$ then ${\rm Sym}^2({\bf C}^m)$ is an even-dimensional orthogonal representation of ${\rm SO}_m$ each of whose elements has a $1$-eigenvector). $\endgroup$
    – Noam D. Elkies
    Commented Mar 11, 2015 at 3:34
  • $\begingroup$ If $G$ is connected, if we consider an invariant flag $(V_i)$ of subspaces of $V=\mathbf{C}^n$ on which $G$ acts irreducibly on every $V_{i+1}/V_i$ then there exists $i$ such that every element of $G$ has eigenvalue 1 on $V_{i+1}/V_i$. So things boils down (for connected $G$) to the case when $G$ is reductive and acts irreducibly (take the Zariski closure if necessary), and actually semisimple (because the center acts by scalar matrices, hence is trivial). $\endgroup$
    – YCor
    Commented Mar 11, 2015 at 21:53

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