We know that for any $\beta \in \mathbb{R}$, the following equality holds: $$ \mathbb{E}\left[\exp(\beta X_t)\right]=\exp\left(\mu \beta t+\frac{1}{2} \sigma^2 \beta^2 t\right) $$ Can we conclude that $X_t \sim \mathcal{N}(\mu t,\sigma^2 t)$?
I notice that the moment generating function uniquely determines the distribution. Hence, we can get the desired result directly.