I'm currently reading through some material on vector fields and came across the remark that the vector field $\frac{d}{dt}$ does not form any one-parameter subgroups on $(0,1)$.
This seems quite natural as the one-parameter subgroup generated by $\frac{d}{dt}$ is the translation map $\varphi_h:t\mapsto t+h$ which is not a diffeomorphism on $(0,1)$. However, the remark comes after the proof that a vector field of compact support always has a one-parameter subgroup. This seems to suggest that any vector field on a compact manifold should generate one-parameter subgroups.
With this said, it seems that $[0,1]$ should have a well-defined one-parameter subgroup corresponding to $\frac{d}{dt}$.
Is this the case? It seems to me that $[0,1]$ should fail to have a one-parameter subgroup for the same reason as $(0,1)$.
The result is that any vector field on a compact manifold without boundary generates a one-parameter group. $[0,1]$ is not a manifold without boundary and, indeed, the vector field $\frac{d}{dt}$ does not generate a one-parameter group for the reason you describe.