Self-similarity of Brownian motion

Question

I learned in class that Brownian motion is a 0.5-self similar process, which implies that $$(T^{0.5}B_{t_1},\dots,T^{0.5}B_{t_n})\sim (B_{Tt_1},\dots, B_{Tt_n})$$ for any $T>0$. I am solving a problem and I got $\mathbb{E}(e^{\sigma B_t})$. Does self-similarity imply that $$\mathbb{E}(e^{\sigma B_t})=\mathbb{E}(e^{B_{\sigma^2t}})?$$