If $(N_t)_{t \geq 0}$ is a Lévy process (i.e. a stochastic process with stationary and independent increments) and $h \neq 0$, then the random variables $N_t$ and $N_{t+h}$ cannot have the same distribution unless $(N_t)_{t \geq 0}$ is trivial, i.e. $N_t=0$ for all $t \geq 0$. This implies, in particular, that the only stationary Lévy process is the trivial process.
To prove this, assume that $(N_t)_{t \geq 0}$ is a non-trivial Lévy process and fix $t,h \geq 0$ such that $N_t$ and $N_{t+h}$ have the same distribution. Since $$N_{t+h} = (N_{t+h}-N_t) + N_t$$ it follows from the independence and stationarity of the increments that $$\mathbb{E}\exp(i \xi N_{t+h}) = \mathbb{E}\exp(i \xi N_t) \exp(i \xi N_h)$$ for all $\xi$. By assumption, $N_t$ and $N_{t+h}$ have the same distribution, and so $$\mathbb{E}\exp(i \xi N_{t}) = \mathbb{E}\exp(i \xi N_t) \exp(i \xi N_h)$$ for all $\xi$. Since $\mathbb{E}\exp(i \xi N_t) \neq 0$ for all $\xi$ (that's a property of Lévy processes) it follows that $$1= \mathbb{E}\exp(i \xi N_h)$$ for all $\xi$. Note that $\chi(\xi)=1$ is the Fouier transform of the random variable $X=0$. By the uniqueness of the Fourier transform, we therefore get $N_h=0$ almost surely. This is, however, only possible if $h=0$; this can be, for instance, proved using the Lévy-Khintchine formula. Hence, we conclude that $N_t$ and $N_{t+h}$ can only have the same distribution in the trivial cases that $h=0$.
