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Noether's theorem states that every differentiable symmetry of the action of a physical system has a corresponding conservation law.

Suppose our action is of the form $S = \int d^4x\, \mathcal{L}(\phi,\partial_\mu\phi).\tag{1}$

if $x \rightarrow x'$ then if $S \rightarrow S'$ where
$S' = \int d^4x'\, \mathcal{L'}(\phi',\partial_\mu\phi').\tag{2}$

But from calculus we know that $S=S'$ so does that mean that every change of variable correspond to a conserved quantity? why the quantities conserved under Poincare transformation, for example, is more especial?

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The action shown in the question is a functional of $\phi$, not of $x$. A change of the integration variable $x$ is just a relabeling of the index set. It does not transform the dynamic variables $\phi$ at all, so no: a change of variable does not correspond to a conserved quantity.

More explicitly, if $y(x)$ is a monotonic smooth function of $x$, then $$ \int d^4y\ {\cal L}\left(\phi\big(y(x)\big),\, \frac{\partial}{\partial y_\mu}\phi\big(y(x)\big)\right) = \int d^4x\ {\cal L}\left(\phi(x),\frac{\partial}{\partial x_\mu}\phi(x)\right) \tag{1} $$ identically, for any ${\cal L}$ whatsoever (as long as it depends on $x$ only via $\phi$). This is just a change of variable (a relabeling of the index-set), and there is no associated conserved quantity.

In contrast, suppose that the action has this property: $$ \int d^4x\ {\cal L}\left(\phi\big(y(x)\big),\, \frac{\partial}{\partial x_\mu}\phi\big(y(x)\big)\right) = \int d^4x\ {\cal L}\left(\phi(x),\frac{\partial}{\partial x_\mu}\phi(x)\right). \tag{2} $$ Unlike equation (1), equation (2) is not identically true for any ${\cal L}$ and any $y(x)$, though it may be true for some choices of ${\cal L}$ and $y(x)$. The transformation represented in equation (2) replaces the original function $x$, namely $\phi(x)$, with a new function of $x$, namely $\phi\big(y(x)\big)$. This is the kind of transformation we have in mind when we talk about Poincaré invariance and its associated conserved quantities: it is a change of the function $\phi$ which we then insert into the original action, not a change of the integration variable.

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  1. Note that in the context of Noether's theorem that some quasisymmetry transformations are so-called trivial quasisymmetry transformations, and corresponds to a trivial conservation law, cf. e.g. my Phys.SE answer here.

  2. In particular, if we consider a passive coordinate transformation of the system, then the action $S$ is trivially invariant, and we can e.g. use the trick with $x$-dependent infinitesimal $\epsilon(x)$ to conclude that the corresponding full Noether current $J^{\mu}$ vanishes identically.

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