Let $E$ be a Banach space, and let $(\sigma_t)$ be a strongly continuous one-parameter group on $E$: so for $t\in\mathbb R$, we have that $\sigma_t$ is a contraction on $E$, $\sigma_t \sigma_s=\sigma_{t+s}$, $\sigma_0$ is the identity, and for each $x\in E$, the map $\mathbb R\rightarrow E; t\mapsto\sigma_t(x)$ is continuous.

We have the notion of the analytic continuation of $(\sigma_t)$. For example, set $S(i) = \{ z=t+iy\in\mathbb C : 0<y<1 \}$ with closure $\overline{S(i)}$. Then $D(\sigma_i)$ consists of those $x\in E$ such that there is a bounded continuous map $f:\overline{S(i)}\rightarrow E$ with $f$ analytic on $S(i)$, and with $f(t) = \sigma_t(x)$ for $t\in\mathbb R$. Then we set $\sigma_i(x) = f(i)$. Then $\sigma_i$ is a densely defined, closed operator.

Under favourable circumstances, the following is somehow "folklore":

Let $x\in D(\sigma_i)$ with $\sigma_i(x)=x$. Then $\sigma_t(x)=x$ for all $t$.

E.g. if E is a Hilbert space this can be extracted from Stone's Theorem and some spectral theory. What I want to know is if there is a way to prove this using the more abstract framework I have setup (which is more accessible if, e.g. we start considering dual spaces and weakly-continuous groups etc.)