= Solution
Integrating the vector-field equations gives the one-parameter transformations
$$
\boxed{(x,t,v)\longmapsto(e^{\alpha s}x,e^{\beta s}t,e^{\gamma s}v).}
$$
Under this scaling, $v_t,v^2v_x,v_{xxx}$ have weights $\gamma-\beta,3\gamma-\alpha,\gamma-3\alpha$. Equality of all three weights gives
$$
\boxed{\beta=3\alpha,\qquad\gamma=-\alpha.}
$$
Thus the nontrivial <Lie point symmetry> generator can be normalized to $x\partial_x+3t\partial_t-v\partial_v$.
For the <Miura transformation> $u=v^2+v_x$, direct <differentiation> gives the operator identity
$$
u_t-6uu_x+u_{xxx}
=(\partial_x+2v)(v_t-6v^2v_x+v_{xxx}).
$$
For example $u_x=2vv_x+v_{xx}$ and $u_{xxx}=6v_xv_{xx}+2vv_{xxx}+v_{xxxx}$; inserting them makes the $v_xv_{xx}$ terms cancel and leaves the right side. Hence \b[$u$ solves $u_t-6uu_x+u_{xxx}=0$] whenever $v$ solves the <mKdV equation>. Both $v^2$ and $v_x$ have weight $-2\alpha$, so the corresponding <Korteweg-De Vries equation> symmetry is
$$
\boxed{(x,t,u)\mapsto(e^s x,e^{3s}t,e^{-2s}u),\qquad
x\partial_x+3t\partial_t-2u\partial_u.}
$$
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