Conserved quantity of a relativistic free Lagrangian for a Lorentz boost Let 
$$L~=~-mc^2\sqrt{1- \frac{|\textbf{v}|^2}{c^2} },$$ 
where $\textbf{v}$ is the usual velocity of the particle in a fixed inertial frame. Then, this is the Lagrangian for a relativistic free particle. Now what does it mean by "the conserved quantity for a Lorentz boost"? Does it mean that the particle is boosted by some fixed velocity and there comes out a quantity that is preserved? I cannot get the exact meaning of the phrase. Could anyone please explain to me?
 A: This is easiest to see from the Hamiltonian formulation, cf. this Phys.SE post. Below we will also give the non-relativistic expressions for comparison, because it is interesting and somewhat subtle, cf. this Phys.SE post, and because OP asked the analogous non-relativistic question earlier.
I) The Hamiltonian is the kinetic energy, i.e the energy minus the rest energy$^1$
$$\begin{align}  H~=~&p^0c-mc^2~=~\sqrt{{\bf p}^2c^2+m^2c^4}-mc^2\cr
\longrightarrow&\quad
\frac{{\bf p}^2}{2m}\quad\text{for}\quad c\to \infty.\end{align} \tag{1} $$
The 3 boost generators $B^i$ are part of the 6 Lorentz generators
$$\begin{align}B^i~=~&\frac{J^{0i}}{c} ~=~tp^i-x^i\frac{p^0}{c}\cr 
\longrightarrow &\quad 
tp^i-mx^i\quad\text{for}\quad c\to \infty,\cr 
& i~\in~\{1,2,3\}.\end{align}\tag{2}$$
The pertinent infinitesimal quasi-symmetry transformations are generated by the boosts
$$\begin{align}\delta x^i ~=~&\{ x^i , {\bf B}\cdot \delta {\bf v} \}
~=~t~\delta  v^i - \frac{p^i}{p^0c}{\bf x}\cdot \delta {\bf v}\cr
\longrightarrow &\quad t~\delta  v^i\quad\text{for}\quad c\to \infty, \tag{3}\cr 
\delta p^i ~=~&\{ x^i , {\bf B}\cdot \delta {\bf v} \}
~=~\frac{p^0}{c}\delta  v^i\cr
\longrightarrow &\quad 0\quad\text{for}\quad c\to \infty,\tag{4}\cr
 \delta t~=~&0.\tag{5}\end{align}$$
The Hamiltonian Lagrangian
$$\begin{align} L_H ~=~&{\bf p}\cdot \dot{\bf x}  - H \cr
\longrightarrow &\quad  
{\bf p}\cdot \dot{\bf x}  - \frac{{\bf p}^2}{2m} \quad\text{for}\quad c\to \infty \end{align}\tag{6}$$
has a quasi-symmetry
$$\begin{align}\delta L_H~=~&\frac{d}{dt}\left( 
\frac{m^2c}{p^0} {\bf x}\cdot \delta {\bf v} \right)\cr
\longrightarrow &\quad \frac{d}{dt}\left( 
m {\bf x}\cdot \delta {\bf v} \right) \quad\text{for}\quad c\to \infty. \end{align}\tag{7}$$
One may check that the corresponding Noether charges are precisely the boost generators (2).
II) The corresponding Lagrangian formulation$^1$
$$\begin{align}L~=~&mc^2\left(1-\sqrt{1-\frac{\dot{\bf x}^2}{c^2}}\right)\cr
\longrightarrow &\quad \frac{1}{2}m\dot{\bf x}^2\quad\text{for}\quad c\to \infty,\end{align}\tag{8}$$
has infinitesimal boost quasi-symmetry
$$\begin{align}\delta x^i 
~=~&t~\delta  v^i - \frac{\dot{x}^i}{c^2}{\bf x}\cdot \delta {\bf v}\cr
\longrightarrow &\quad t~\delta  v^i\quad\text{for}\quad c\to \infty, \tag{9}\cr
\delta t~=~&0,\tag{10}\end{align}$$
and conserved boost charges
$$\begin{align}B^i~=~&m\frac{t\dot{x}^i-x^i}{\sqrt{1-\frac{\dot{\bf x}^2}{c^2}}} \cr
\longrightarrow &\quad 
m(t\dot{x}^i-x^i)\quad\text{for}\quad c\to \infty.\tag{11}\end{align}$$
This is left as an exercise to the reader. One way is to integrate out the 3-momentum ${\bf p}$ from the Hamiltonian formulation in section I. See also this related Phys.SE posts and links therein.
--
$^1$ We have removed the rest energy, which is a constant, i.e. a total time derivative, in order to be able to go to the non-relativistic limit.
