"Deriving" Newton's laws of motion from symmetry assumptions It is often discussed how certain symmetries and conservation laws can be derived from Newton's laws of motion. My question is: can we go the other way? Can Newton's laws of motion be derived only from symmetry assumptions on a dynamical system such as time reversal, time translation, space translation, etc.?
There are clearly more assumptions and definitions needed, e.g. what exactly is meant by a dynamical system (perhaps a state in $\mathbb{R}^n$ and a differential equation stating the dynamics). However, I think there is no need for more "arbitrary" assumptions.
I believe that specifying the symmetries of a system is a more general and natural way of expressing assumptions about it. Practically, it may help with linking classical Newtonian mechanics to dynamical systems that have similar symmetries. Philosophically, it might also illustrate how "arbitrary" these laws are.
 A: Yes, there is a very important theorem in physics called Noether's theorem (or more specifically Noether's 1st theorem) which definitively links symmetries and conservation laws. It states that for every continuous symmetry of the action of a system, there exists a corresponding conserved current. In simpler terms, if the action, which is an integral of the Lagrangian over all possible paths, is invariant under a symmetry group with continuous parameters, then you'll get out a conservation law for that system.
It turns out that time translation symmetry gives energy conservation and spatial translation symmetry gives conservation of momentum.  Additionally, for example, rotational invariance gives rise to conservation of angular momentum.
Furthermore, following the formalism of Lagrangian mechanics, you can use the Lagrangian of your system to derive the system's Newtonian equations of motion. Start with a Lagrangian: $$L(q, \dot{q}, t) = T -V$$
Then construct the Euler-Lagrange equation(s): $$\frac{d}{dt}\big (\frac{\partial L}{\partial \dot{q}} \big ) - \frac{\partial L}{\partial q} = 0.$$
A: In this answer I discuss the following two things:
-Among the multiple symmetries of mechanics is the symmetrical relation between the members of the equivalence class of inertial coordinate systems. (I categorize the property of reciprocity as a form of symmetry.)
-Symmetry properties of F=ma

The equivalance of the set of inertial coordinate systems has been treated by many; I find the treatment by Palash B. Pal particularly effective. The title of the article is 'Nothing but Relativity' Palash B. Pal demonstrates that the constraint of symmetrical relation between the members of the equivalence class of inertial coordinate system is sufficient to narrow down the options to just two: the Galilean transformations and the Lorentz transformations.

As we know: newtonian mechanics is a limiting case of relativistic mechanics; for non-relativistic velocity the expressions simplify to the expressions for newtonian mechanics..
Part of the set properties that must be granted in order to have special relativity is granting Minkowski spacetime.
The corresponding assumption for newtonian mechanics is the assumption that space-and-time are Euclidean. This includes granting that space and time stand in a particular relation to each other: for newtonian mechanics it must not only be granted that an object released to free motion will move along a straight line; it must also be granted that the object will in equal time intervals cover equal intervals of space.
(Historically its counterpart for the motion of an object being subjected to a central force was formulated earlier: an object in central-force-caused circumnavigating motion will in equal intervals of time sweep out equal areas; Kepler's second law.)

The set of relations between symmetries and corresponding conserved quantities is commonly described using Noether's theorem. It seems to me however, that those relations are not dependent on describing them with Noether's theorem. That is: I don't think Noether's theorem is a necessity to describe those correspondencies, it just so happens that historically that was the mathematical tool involved in the first recognition of those correspondencies.


Symmetry properties of F = ma
Using the standard letters for time, position, velocity and acceleration:
t time
s position
v velocity
a acceleration
I regard the following as an instance of symmetry:
$$ v = \frac{ds}{dt}   \tag 1 $$
$$ a = \frac{dv}{dt}   \tag 2 $$
Velocity is to position is what acceleration is to velocity.

That symmetry goes a long way.
I start with the case of uniform acceleration:
$$ v = at \tag 3 $$
To find distance covered as a function of time: integrate (3) with respect to the time coordinate:
$$ s = \int_0^t at \ dt =  \tfrac{1}{2}at^2  $$
$$ s = \tfrac{1}{2}at^2    \tag 4 $$
Combining the relations (3) and (4) is in effect combining the relations (1) and (2):
(5) is obtained by multiplying both sides of (4) with acceleration $a$, and then applying (3):
$$ as = a\tfrac{1}{2}at^2 = \tfrac{1}{2}a^2t^2 = \tfrac{1}{2}(at)^2 = \tfrac{1}{2}v^2     $$
$$ as = \tfrac{1}{2}v^2     \tag 5 $$
The product $as$ skips the entity in the middle: the velocity $v$.
In (5) the product $as$ transforms to the square $v^2$ by transfering a factor $t$ (time)
The position coordinate $s$ is rescaled to velocity by dividing it with the time factor $t$, and the acceleration value is rescaled to velocity by multiplying it with the time factor $t$.

Generalize (5) to accommodate an initial velocity $v_0$, and an initial position $s_0$:
$$ a(s -s_0) = \tfrac{1}{2}v^2 - \tfrac{1}{2}{v_0}^2     \tag 6 $$

For non-uniform acceleration: integration.
The relations (1) and (2) are then used in the following form:
$$ v = \frac{ds}{dt} \qquad \Longleftrightarrow \qquad ds = v \ dt  \tag 7 $$
$$ a = \frac{dv}{dt} \qquad \Longleftrightarrow \qquad dv = a \ dt  \tag 8 $$
The integral:
$$ \int_{s_0}^s a \ ds = \int_{t_0}^t a \ v \ dt = \int_{t_0}^t v \ a \ dt = \int_{v_0}^v v \ dv = \tfrac{1}{2}v^2 - \tfrac{1}{2}{v_0}^2   $$
$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}{v_0}^2     \tag {9} $$
As we know: integration is addition of rectangular strips.
The integral $\int_{s_0}^s a \ ds$ specifies rectangular strips with width $ds$ and height $a$.
The integral $\int_{v_0}^v v \ dv$ specifies rectangular strips with width $dv$ and height $v$.
The dimensions of the rectangular strips are rescaled; the height is multiplied with the factor $t$ and the width is divided by the factor $t$, that is why the areas are the same.

All of the above is obtained by virtue of the symmetry:
$$ v = \frac{ds}{dt}   \tag 1 $$
$$ a = \frac{dv}{dt}   \tag 2 $$


We obtain the work-energy theorem by combining $F=ma$ with (9):
$$ \int_{s_0}^s F \ ds = \int_{s_0}^s ma \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}m{v_0}^2    $$
$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}m{v_0}^2     \tag {10} $$


Historically: the concepts of potential energy and kinetic energy were in circulation before there the work-energy theorem was stated in the modern form. The work-energy theorem clarifies what the concept of energy is: it shows that potential energy and kinetic energy have in common that they both arise from integration with respect to the position coordinate.
For the expression for kinetic energy $ \tfrac{1}{2}mv^2$: diffentiation with respect to the position coordinate recovers $ma$:
$$ \frac{d(\tfrac{1}{2}mv^2)}{ds} = \tfrac{1}{2}m\left( 2v\frac{dv}{ds} \right) = m\frac{ds}{dt}\frac{dv}{ds} = m\frac{dv}{dt} = ma  $$
$$ \frac{d(\tfrac{1}{2}mv^2)}{ds} = ma  \tag {11} $$
And of course: when you differentiate potential energy with respect to the position coordinate you recover the force.
The definition of potential energy takes the content of the work-energy theorem to advantage; potential energy is defined such that over time the value of the sum $\Delta E_p$ and $\Delta E_k$ is constant:
$$ \Delta E_p = - \int_{s_0}^s F \ ds \tag {12} $$
$$ - \Delta E_p = \Delta E_k \tag {13} $$
$$ \Delta E_k + \Delta E_p = 0 \tag {14} $$
$$ \frac{d(E_k + E_p)}{dt} = 0 \tag {15} $$
(15) expresses explicitly the symmetry under time translation: if the derivative with respect to time is zero then as you change the time coordinate the value of $(E_k + E_p)$ remains the same.

Going back once more to (1) and (2): I regard that symmetry as the origin of (14)(15).
$$ v = \frac{ds}{dt}   \tag 1 $$
$$ a = \frac{dv}{dt}   \tag 2 $$
A: QFT uses a similar symmetry-to-laws approach I'll adapt to your question. The goal is to build the most general Lagrangian compatible with the desired symmetries and some other conditions, such as (i) smooth dependence on phase space coordinates, which demands a quadratic kinetic sector and (ii) not going further than first-order derivatives because they cause a problem we'll try to avoid. Assuming Lagrangians describe the physics might already feel like a cheat, but they're our clearest understanding of how to start from symmetries, so we'll settle for seeing how symmetries motivate a specific Lagrangian from which Newton's second law, in particular, emerges.
Whereas in the QFT of a Lorentz scalar $\phi$ this gives a Lagrangian density of the form $\mathcal{L}=\frac12\partial_\mu\phi\partial^\mu\phi-V(\phi)$, the problem at hand obtains a true Lagrangian because we only want to work with time-dependent particle positions, not spacetime-dependent fields. (The latter don't originate with QFT or even QM, as e.g. electromagnetism or fluid mechanics illustrate.) Hence we get $L=\frac12m\dot{x}^2-V(\vec{x})$, where we've scaled $\vec{x}$ so a mass factor $m$ can appear in the kinetic term. This allows us to easily upgrade to an arbitrarily diverse universe of particles, viz. $L=\frac12\sum_{i=1}^nm_i\dot{x}_i^2-V(\vec{x}_1,\,\cdots,\,\vec{x}_n)$. Our Euler–Lagrange equations are now $\frac{d\vec{p}_i}{dt}=-\nabla_{\vec{x}_i}V$ for $\vec{p}_i:=m_i\dot{\vec{x}}_i$.
Next we need to be careful about an issue that's often much harder to explain from a Lagrangian perspective than a Newtonian one: how do we accommodate non-conservative forces? Short answer, like this; long answer, to help with our present context, the symmetry assumption from which we start has to be weakened from $\frac{\delta S}{\delta\vec{x}_i}=\vec{0}$ to $\frac{\delta S}{\delta\vec{x}_i}=-\vec{Q}_i^p$. This allows us to separate the question of which symmetries appear on the left-hand side from the question of which non-conservative forces appear on the right. In either case,$$\frac{\delta S}{\delta\vec{x}_i}:=\frac{\partial L}{\partial\vec{x}_i}-\frac{d}{dt}\frac{\partial L}{\partial\dot{\vec{x}}_i}.$$This separation again may bother you, but bear in mind the dissipative version of Newton's second law is similar: $\frac{d\vec{p}_i}{dt}=-\nabla_{\vec{x}_i}V-\vec{Q}_i^p$. What's more, dissipative forces are a symptom of pour choosing to analyze systems that are physically incomplete, thereby susceptible to something we didn't model. For example, if you describe friction in terms of individual atoms, you no longer need to describe one macroscopic object losing net energy. Unsurprisingly, QFT is so fundamental it obtains conservative Lagrangian densities. (You can invent a dissipative field theory, then quantize its fields, but the real world doesn't seem to work that way.)
We still haven't identified how $\vec{x}_i$ transforms under a different choice of coordinates. The answer must emerge from demanding our ELEs are thereby invariant. So we want to derive Galilean transformations from Newton's second law. That's a subtle issue discussed here.
