Let $(V,\langle.,.\rangle)$ be a Euclidean vector space defined over $\mathbb{R}$ of odd dimension $n $ and let $f : V \rightarrow V$ an orthogonal mapping with $\operatorname{det}(f)=1 $. Then the map $ f $ has a vector $v \in V \backslash\{0\} $ with $ f(v)=v $.
Now this is obviously true since $ f$ is orthogonal it preserves distances: $||f(v)|| = ||v|| \Rightarrow f(v) = \lambda v \Rightarrow |\lambda| = 1$.
This should be true for even dimensions as well because $||f(v)|| = ||v||$ should hold for every finite dimensional $\mathbb{R}$ vector space. Is this the argument here? If there is a more detailed argument, feel free to share!
You showed that all eigenvalues have absolute value one, that is, lie on the unit circle. That does not imply that any of these points is $1+i\,0$.
Remember that in a polynomial with real coefficients, the complex roots come in conjugate pairs.
In $\Bbb R^4$ you can combine a rotation in the first two coordinates with an independent rotation in the second two variables.