I have a compact self-adjoint positive integral operator $Q:L^2(0, \infty) \to L^2(0, \infty)$ with operator norm $\| Q\| =1$. By the assumptions, we know $1$ is an eigenvalue of $Q$. Let $y\ne 0$ and consider applying a rotation after $Q$: $$ T_y[f](t) = e^{iyt}Q[f](t). $$
It is not hard to see $\|T_y\|=1$. Is it true that $1$ cannot be an eigenvalue of $T_y$?
I think the answer is false in finite dimensional spaces, because we can pick for example rotation by a full circle.
Attempt: I could only deal with $|y|$ small. Since $T_y$ is an analytic perturbation of the operator $Q$, by Kato-Rellich Theorem, there is an analytic function $y\to \lambda(y)$ with $\lambda(0)=1$ (eigenvalue is analytic). By uniqueness of analytic continuation, the set $\{ y\mid \lambda(y)=1 \}$ must not accumulate. It follows that for small nonzero $y$, $\lambda(y)\ne 1$.
If you are intersted, my operator is $$ Q[f](t) = \int_0^\infty f(s) e^{-(t^2+s^2) - \beta(t+s)} I_0(2\sqrt{ts}) ds $$ with $\beta$ suitably chosen so that $\|Q\|=1$.
Assume $T_yf=f.$ Thus $(Qf)(t)=e^{-iyt}f(t).$ Then $$0\le \langle Qf,f\rangle =\int\limits_0^\infty e^{-iyt}|f(t)|^2\,dt$$ For every positive operator we have $$\|Qf\|^2\le \|Q\|\,\langle Qf,f\rangle$$ Therefore $$\int\limits_0^\infty |f(t)|^2\,dt \le \int\limits_0^\infty e^{-iyt}|f(t)|^2\,dt$$ i.e. $$\int\limits_0^\infty [\cos yt-1]\,f(t)|^2\,dt\ge 0$$ Thus $[\cos yt-1]f(t)=0,\ a.e.$ hence $f=0.$