I feel like there must be a way to deduce the (special) relativstic Lagrangian using only the Lorentz transform (without other knowledge of special relativity). So far, all the derivations I have seen say something like
The action must be a Lorentz scalar, so naturally the Lagrangian has to be $$L=-\gamma^{-1} - V$$
(where I use $c=m=1$). I understand why the action must be a Lorentz scalar but I do not understand why this Lagrangian is the only one possible. For the past hours I tried to convince myself of that but I am stuck. I feel like I am on the right path and I would much appreciate help. What follows is my reasoning so far.
For now let's work in 1 space and 1 time dimensions.
We want the laws of physics to be the same for different observers related by a Lorentz transform (which I note $T_L(v)$). This means that if in one frame of reference the action, S, is stationnary,
$$ \delta S = 0 $$
then we want it to also be stationnary in another frame of reference
$$ T_L(v) \delta S = 0 $$
The effect of $T_L$ on $S$ is
$$ T_L(v) S(L) = T_L \int_{t_1}^{t_2}dt L(x,\dot{x},t) = \int_{\gamma t_1}^{\gamma t_2}dt \gamma L(T_L(v)\{x,\dot{x},t\}) = \int_{\gamma t_1}^{\gamma t_2} dt \gamma L(\gamma (x-vt), \frac{\dot{x}-v}{1-\dot{x}v},\gamma(t-vx)) $$
where $\gamma = (1-v^2)^{-1/2}$. Let's restore the integration bounds to $t_1,t_2$ such that the effect of the transformation can be fully written in terms of $L$ (hopefully it will be clearer what I mean in what follows). Let's substitute $u=\gamma^{-1} t$:
$$ T_L(v)S(L) = \int_{t_1}^{t_2}du \gamma^2 L(\gamma (x-\gamma u v), \frac{\frac{1}{\gamma}\frac{d x}{du}-v}{1-\frac{1}{\gamma}\frac{dx}{du}v}, \gamma(\gamma u - vx) )$$
Now, u is being integrated over and plays the same role as t played previously, we can thus write
$$ T_L(v)S(L) = S(\gamma^2 L(\gamma (x-\gamma u v), \frac{\frac{1}{\gamma}\frac{d x}{du}-v}{1-\frac{1}{\gamma}\frac{dx}{du}v}, \gamma(\gamma u - vx) ) $$
Using the action, S, as a basis, we can write that as
$$ T_L(v)S=e^{ivX}S $$
since $X=-i\frac{d}{dv}T_L(v)S(L)|_{v=0}$, we can deduce
$$ X = -i[(\frac{d}{dv} (\gamma^2 L(\gamma (x-\gamma u v), \frac{\frac{1}{\gamma}\frac{d x}{du}-v}{1-\frac{1}{\gamma}\frac{dx}{du}v}, \gamma(\gamma u - vx) ))\frac{\delta }{\delta L}]_{v=0} $$
Ugly calculations yield
$$ X = -i (-t \frac{\partial L(x,\dot{x},t)}{\partial x} - x \frac{\partial L(x,\dot{x},t)}{\partial t}-(1-\dot{x}^2)\frac{\partial L(x,\dot{x},t)}{\partial \dot{x}})\frac{\delta}{\delta L} $$
If we want the action to not change, then we need $e^{ivX} = 1 \forall v$, thus
$$ t \frac{\partial L(x,\dot{x},t)}{\partial x} + x \frac{\partial L(x,\dot{x},t)}{\partial t}+(1-\dot{x}^2)\frac{\partial L(x,\dot{x},t)}{\partial \dot{x}} = 0$$
If we want the action to not change around the minimum (or rather stationnary point), then we can invoke the Euler-Lagrange equation which gives
$$ \frac{d}{dt} (t \frac{\partial L}{\partial \dot{x}})+x\frac{\partial L}{\partial t}-\dot{x}^2\frac{\partial L}{\partial \dot{x}}=0$$
Either way, I am stuck here and just plugging in the known result for a free particle, $L = \sqrt{1-\dot{x}^2}$ gives
$$ -\dot{x} L = t \frac{\partial L}{\partial x} + x \frac{\partial L}{\partial t} = 0$$
which is obviously nonsense. Where did I go wrong?
