I am not getting any clue to show that $t \to e^{it}$ and $t \to e^{2it}$ are not equivalent as curves. But we get the same picture when we draw but the speed by which we travel is 2 times in the second than the first. By equivalent we mean that there is a continuous bijection.
The map is from the set $[0,2\pi]$.
A curve is said to be equivalent to another curve if there exists a smooth bijection from the domain of the first curve to the domain of the second curve.