Also Qmechanic gave the correct answer, I believe it is overloaded, because there is actually no need to use motion equations (Euler-Lagrange equations) to answer to the second part of the OP question at least.

Actually you can simply just generalize original Landau approach to this issue for an answer, so I will mention it here in details:

Lets suppose that $L\left(\vec{v}^{2n}\right)$($2n$ is to have scalar value) is the Lagrangian of a free particle in inertial frame $K$. suppose another inertial frames of reference $K'$ that moves relative to $K$ with infinitesimal velocity $\vec{\varepsilon}$, the Lagrangian $L'=L\left[\left(\vec{v}+\vec{\varepsilon}\right)^{2n}\right]$ in $K'$ that describes the particle should be same Lagrangian as in $K$ up to a total time derivative.

To show that, we expand $\left(\vec{v}+\vec{\varepsilon}\right)^{2n}$ in first order of $\vec{\varepsilon}$, to find it, we suppose at first that $n=1$, then: $$\left(\vec{v}+\vec{\varepsilon}\right)^{2}\simeq v^{2}+2\vec{\varepsilon}\cdot\vec{v}$$ then to find for $n=2$ we write: $$\left(\vec{v}+\vec{\varepsilon}\right)^{4}\simeq\left(v^{2}+2\vec{\varepsilon}\cdot\vec{v}\right)\left(v^{2}+2\vec{\varepsilon}\cdot\vec{v}\right)\simeq v^{4}+4\left(\vec{\varepsilon}\cdot\vec{v}\right)v^{2}$$ repeating this couple times insures you that: $$\left(\vec{v}+\vec{\varepsilon}\right)^{2n}=v^{2n}+2n\left(\vec{\varepsilon}\cdot\vec{v}\right)v^{2n-2}+O\left(\varepsilon^{2}\right)$$ then we can write by expanding the Lagrangian that: $$L\left[\left(\vec{v}+\vec{\varepsilon}\right)^{2n}\right]\simeq L\left(\vec{v}^{2n}\right)+2n\left(\vec{\varepsilon}\cdot\vec{v}\right)v^{2n-2}\frac{\partial L}{\partial\vec{v}^{2n}}\simeq L\left(v^{2n}\right)+2n\, v^{2n-2}\frac{\partial L}{\partial v^{2n}}\sum_{i}\varepsilon_{i}v_{i}$$

As a special case, just to make things clear, we see that for $n=1$ we get: $$L\left[\left(\vec{v}+\vec{\varepsilon}\right)^{2}\right]\simeq L\left(v^{2}\right)+2\frac{\partial L}{\partial v^{2}}\sum_{i}\varepsilon_{i}\frac{dx_{i}}{dt}$$ then we see that to have second term as a full time derivative, the only option for us (putting in mind that Lagrangian is only a function of speed, not velocity) is to have: $$\partial L/\partial v^{2}=const\Rightarrow L=\alpha v^{2}+\beta$$

Back to general case, we see, that what ever $L$ actually is, we have: $$v^{2n-2}\frac{\partial L}{\partial v^{2n}}=g\left(\left\Vert \vec{v}\right\Vert \right)$$ (where $g$ is some function) that is, this term can be only a variable of the speed (not velocity components), so to have out of it total time derivative, again we need to have $g\left(v\right)=const$, this requires that $n=1$ (note that $n>0$), and then Lagrangian form is same as above mentioned special case.

Also Qmechanic gave the correct answer, I believe it is overloaded, because there is actually no need to use motion equations (Euler-Lagrange equations) to answer to the second part of the OP question at least.

Actually you can simply just generalize original Landau approach to this issue for an answer, so I will mention it here in details:

Lets suppose that $L\left(\vec{v}^{2n}\right)$($2n$ is to have scalar value) is the Lagrangian of a free particle in inertial frame $K$. suppose another inertial frames of reference $K'$ that moves relative to $K$ with infinitesimal velocity $\vec{\varepsilon}$, the Lagrangian $L'=L\left[\left(\vec{v}+\vec{\varepsilon}\right)^{2n}\right]$ in $K'$ that describes the particle should be same Lagrangian as in $K$ up to a total time derivative.

To show that, we expand $\left(\vec{v}+\vec{\varepsilon}\right)^{2n}$ in first order of $\vec{\varepsilon}$, to find it, we suppose at first that $n=1$, then: $$\left(\vec{v}+\vec{\varepsilon}\right)^{2}\simeq v^{2}+2\vec{\varepsilon}\cdot\vec{v}$$ then to find for $n=2$ we write: $$\left(\vec{v}+\vec{\varepsilon}\right)^{4}\simeq\left(v^{2}+2\vec{\varepsilon}\cdot\vec{v}\right)\left(v^{2}+2\vec{\varepsilon}\cdot\vec{v}\right)\simeq v^{4}+4\left(\vec{\varepsilon}\cdot\vec{v}\right)v^{2}$$ repeating this couple times insures you that: $$\left(\vec{v}+\vec{\varepsilon}\right)^{2n}=v^{2n}+2n\left(\vec{\varepsilon}\cdot\vec{v}\right)v^{2n-2}+O\left(\varepsilon^{2}\right)$$ then we can write by expanding the Lagrangian that: $$L\left[\left(\vec{v}+\vec{\varepsilon}\right)^{2n}\right] = L\left(\vec{v}^{2n}\right)+2n\left(\vec{\varepsilon}\cdot\vec{v}\right)v^{2n-2}\frac{\partial L}{\partial\vec{v}^{2n}}+O\left(\varepsilon^{2}\right) \simeq L+g\left(v\right)\sum_{i}\varepsilon_{i}\frac{dx_{i}}{dt}$$ Where: $$g\left(\left\Vert \vec{v}\right\Vert \right)\equiv2n\, v^{2n-2}\frac{\partial L}{\partial v^{2n}}$$ because $L\left(v\right)$, it should be clear that $g$ have to be a function of speed only (not of velocity or it's components), also we see that the sum sign, is actually already a full time derivative by it's own, so to keep the second term full time derivative, we see that the only possible option for us it to have $g\left(v\right)=const$, from this it follows immediately that $n=1$ (note that $n>0$) and $L=\alpha v^{2}+\beta$.