If I have a theory defined on some manifold, my understanding is that the dynamical objects in the theory should carry a representation of the isometry group of that manifold. Moreover, the action $S$ of my theory should be constructed out of combinations of these dynamical objects so that it is invariant under the isometry group.
For example, in a flat-space field theory for a free field transforming as a scalar under the Poincare group, the action is
$$S[\phi, \partial_{\mu}\phi] = \int d^4x \, \frac{1}{2} (\partial_{\mu}\phi)^2 - \frac{m}{2}\phi^2.$$
Now when I act by say, the translation part the isometry group, my understanding is that the field transforms like
$$\phi(x) \rightarrow \phi(x + \epsilon) \approx \phi(x) + \epsilon^{\mu}\partial_{\mu}\phi(x), \\ \partial_{\mu}\phi(x) \rightarrow \partial_{\mu}\phi(x + \epsilon) = \partial_{\mu}\phi(x) + \epsilon^{\nu}\partial_{\nu}\partial_{\mu}\phi(x)$$
but that the bounds of the integral and the measure do not change. This is because if they did, I would just be carrying out a coordinate transformation, under which any integral over a manifold is invariant by definition.
Carrying out the above transformation, the change in the action is
$$\delta S = \int d^4x \,\frac{1}{2} \epsilon^{\nu}\partial_{\nu}\left[(\partial_{\mu}\phi^2) - m\phi^2\right],$$
which is just zero if I assume that fields vanish at infinity.
However, if I try and apply the same reasoning to a classical particle, something goes wrong. If I take the position of a particle $x(t)$ to be my dynamical object, then the action is
$$S[x] = \int_{t_1}^{t_2}dt \, \frac{1}{2} x'^2,$$
and the manifold that I'm integrating over, namely a line segment, doesn't even have an isometry group. Moreover, it is unclear to me how one should even apply a time translation. My guess would be to make the transformation
$$x(t) \rightarrow x(t + \epsilon) \approx x(t) + \epsilon x'(t),$$
while again leaving the measure and bounds untouched. However, then the action changes by
$$\delta S = \int_{t_1}^{t_2} dt\, \frac{\epsilon}{2}\frac{d}{dt}(x')^2,$$
which is not zero.
My questions are as follows:
Am I correct in saying that dynamical objects in a theory should carry a representation of the isometry group of the manifold they are defined on, and that the action should be invariant under the isometry group?
Am I applying isometry transformations correctly? Namely, should I only be transforming the dynamical objects, but not the measure or the bounds on the integral?
If the dynamical objects should carry a representation of the isometry group, how does the classical particle fit into this picture, since a line segment does not have an isometry group?
