Let $M$ be a non-amenable factor. Suppose $\alpha$ is an action of a group $\Bbb R$ on $M$.
Define $$M^{\alpha}:=\{x\in M: \alpha_t(x)=x \text{ for any} \; t\in \Bbb R\}.$$
If we know that there exits $t>0$ such that $\alpha_t=\mathrm{id}$, can we have $M^{\alpha}=\Bbb C 1 $?
No, this is not possible, even without the assumption of non-amenability.
More precisely, the following holds. If $M$ is any von Neumann algebra and $\alpha$ is any action of $\mathbb{R}$ on $M$ such that $M^\alpha = \mathbb{C} 1$ and such that $\alpha_t = \text{id}$ for some $t > 0$, then either $M = \mathbb{C} 1$ or $M \cong L^\infty(\mathbb{R}/a \mathbb{Z})$ for some $a > 0$ with the action $\alpha$ being conjugate to the translation action.
This result is quite classical, see e.g. Corollary 2.4 in https://doi.org/10.1016/0022-1236(74)90019-6
To check that this Corollary 2.4 is indeed applicable, you need the classical theory of spectral subspaces for an action of a compact group. So for completeness, I include the following details, which are all contained in the paper https://doi.org/10.1016/0022-1236(74)90019-6 and a few preceding papers.
Define the closed subgroup $\Gamma \subset \mathbb{R}$ of $t \in \mathbb{R}$ with $\alpha_t = \text{id}$. By assumption, $\Gamma \neq \{0\}$. So either $\Gamma = \mathbb{R}$, in which case $\alpha$ is trivial and $M = M^\alpha = \mathbb{C} 1$, or $\Gamma = a \mathbb{Z}$ for a unique $a > 0$. We may then view $\alpha$ as a faithful ergodic action of the compact abelian group $K=\mathbb{R}/a \mathbb{Z}$ on $M$.
If $\beta$ is any ergodic action of a compact abelian group $K$ on a von Neumann algebra $M$ and if $\omega \in \widehat{K}$ is a character, the spectral subspace $M(\beta,\omega) \subset M$ is defined as the set of elements $x \in M$ satisfying $\beta_k(x) = \omega(k) x$ for all $k \in K$. By ergodicity of the action, $M(\beta,\omega)$ is either $\{0\}$ or consists of the multiples of a unitary element. The spectrum of $\beta$ is defined as the set of $\omega \in \widehat{K}$ for which $M(\beta,\omega)$ is nonzero. Since the product of two unitary eigenvectors is a unitary eigenvector, the spectrum is a subgroup of $\widehat{K}$.
You next define $M_0$ as the von Neumann subalgebra of $M$ generated by all the spectral subspaces. For every $x \in M$ and $\omega \in \widehat{K}$ and integrating w.r.t. the Haar measure, the element $$\int_K \overline{\omega(g)} \, \beta_g(x) \, dg$$ belongs to $M(\beta,\omega)$. Taking linear combinations and limits, it follows that $M_0=M$.
For every $\omega$ in the spectrum of $\beta$, we choose a unitary eigenvector $u(\omega) \in M$ with eigenvalue $\omega$. By ergodicity, $u(\omega_1)u(\omega_2)$ is a multiple of $u(\omega_1 \omega_2)$. So, $u(\omega)$ defines a projective representation of $\widehat{K}$. Since $M_0 = M$, we have that $M$ is generated by the unitaries $u(\omega)$.
At this point, you also get that the spectrum of $\beta$ equals $\Lambda = \widehat{K/K_0}$, where $K_0 \subset K$ is the kernel of $\beta$. So if $\beta$ is a faithful action, the spectrum equals $\widehat{K}$.
Finally, choosing an arbitrary faithful normal state $\varphi_1$ on $M$ and defining $$\varphi(x) = \int_K \varphi_1(\beta_g(x)) \, dg \; ,$$ we get that $\varphi$ is a faithful normal $\beta$-invariant state on $M$. By $\beta$-invariance, we get that $\varphi(x)=0$ whenever $x \in M(\beta,\omega)$ and $\omega \neq 1$. So, $\varphi(u(\omega)) =0$ whenever $\omega \in \Lambda$ and $\omega \neq 1$. It now follows that $\varphi$ is a trace on $M$ and that we have identified $M$ as a twisted group von Neumann algebra $L_\Omega(\Lambda)$, for a $2$-cocycle $\Omega$ on $\Lambda$ with values in $\mathbb{T}$.
If we now go back to our case where $K = \mathbb{R}/a \mathbb{Z}$ and $\alpha$ is faithful, we get that $\widehat{K} = \mathbb{Z}$, which has trivial $2$-cohomology. This then forces $M \cong L(\mathbb{Z})$, which is abelian.
