Transformation has $\delta \mathcal{L}=0$ but $\delta S \neq 0$ and $\delta S \neq \epsilon \int_{\partial\Omega} d^3 x f $ I saw a answer for Why are these two definitions for symmetries in the Lagrangian equivalent?
I have an example has $\delta \mathcal{L}=0$ but $\delta S \neq 0$ and $$\delta S \neq \epsilon \int_{\partial\Omega} d^3 x f .\tag{1} $$
Action
$$S=\int_\Omega d^4x~\mathcal{L},\tag{2}$$
$$\mathcal{L}=\frac{1}{2}(\partial_t \phi(t,\mathbf{x}))^2 -\frac{1}{2}(\partial_\mathbf{x} \phi(t,\mathbf{x}))^2.\tag{3}$$
Transformation:
$$x\rightarrow x'=\lambda x,\tag{4}$$
$$\phi(x)\rightarrow\phi'(x')=\lambda\phi(x).\tag{5}$$
This transformation has 
$$\delta\mathcal{L}=\mathcal{L}(\partial_{x'}\phi'(x'),\phi'(x'))-\mathcal{L}(\partial_{x}\phi(x),\phi(x))=0, \tag{6}$$
$$\delta S= \frac{1}{2}\int_{\Omega'} d^4x' ((\partial_{t'} \phi(t',\mathbf{x'}))^2 -(\partial_{\mathbf{x}'} \phi(t',\mathbf{x}'))^2)
- \frac{1}{2}\int_\Omega d^4x ((\partial_t \phi(t,\mathbf{x}))^2 -(\partial_\mathbf{x} \phi(t,\mathbf{x}))^2)=(\lambda^4-1)S\tag{7}$$
Note: $$\int_{\Omega'}d^4x'=\int_{\Omega}|\frac{\partial x'^\mu}{\partial x^\nu}|d^4x=\int_{\Omega}\lambda^4 d^4 x.\tag{8}$$
My question:
According to this answer, this transformation is L$1$ but is not S$1$ or S$2$. So does this transformation has conserved current? I feel like it's not a symmetry.
PS: According to Qmechanic's answer: there're two different definition of $\delta \mathcal{L}$
The first one:
$$\delta_1 \mathcal{L}=\mathcal{L}(\phi'(x'),\partial'_\mu\phi'(x'),x')-\mathcal{L}(\phi(x ),\partial _\mu\phi (x ),x )= [\mathcal{L}]_\phi \bar\delta \phi+\partial_\mu(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\bar\delta \phi)+ (\partial_\mu \mathcal{L})\delta x^\mu$$
The second one:
$$\delta_2 \mathcal{L}=J \mathcal{L}(\phi'(x'),\partial'_\mu\phi'(x'),x')-\mathcal{L}(\phi(x ),\partial _\mu\phi (x ),x )$$
with Jacobian $J$ 
$$J=\det(\frac{\partial x'^\mu}{\partial x^\nu})$$
$$\delta_2 \mathcal{L}=[\mathcal{L}]_\phi \bar\delta \phi+\partial_\mu(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\bar\delta \phi)+ \partial_\mu (\mathcal{L}\delta x^\mu) = \delta_1 \mathcal{L}+ \mathcal{L} (\partial_\mu \delta x^\mu)$$
When we consider the symmetry transformation, we need to compute $\delta_2 \mathcal{L}$ instead of $\delta_1 \mathcal{L}$. After using the definition of $\delta_2 \mathcal{L}$, the symmetry defined by $\delta_2 \mathcal{L}$ is always consistent to the symmetry defined by $\delta S$.
 A: I) The infinitesimal version of OP's transformation is
$$\begin{align}\text{Total:}&\qquad \phi^{\prime}(x^{\prime})-\phi(x)~=~\delta \phi~=~\varepsilon~\phi, \tag{A}\cr
\text{Horizontal:}&\qquad x^{\prime\mu} -x^{\mu} ~=~\delta x^{\mu}~=~\varepsilon~ x^{\mu}, \tag{B}\cr
\text{Vertical:}&\qquad\phi^{\prime}(x)-\phi(x)~=~  \delta_0 \phi~=~\varepsilon~ (1-x^{\mu}d_{\mu})\phi, \tag{C}
\end{align}$$
where $\varepsilon$ is an infinitesimal parameter. 
II) The action is
$$S[\phi]~=~\int_{\Omega} \mathbb{L}; \tag{D}$$
the Lagrangian 4-form is
$$ \mathbb{L}~=~{\cal L}~ d^4x; \tag{E}$$
and the Lagrangian density is
$$ {\cal L}~=~\frac{1}{2}d_{\mu}\phi ~d^{\mu}\phi. \tag{F} $$
III) The Lagrangian density ${\cal L}$ naively transforms as 
$$ \delta {\cal L}~=~\delta_0 {\cal L}+ \delta x^{\mu} d_{\mu}{\cal L}~=~0 . \qquad\longleftarrow \text{Naive calculation.} \tag{G}$$
However, in case of non-zero horizontal transformations, formula (G) is not a decisive quantity for Noether's theorem, and one should rather look at the action 
$$ \delta S~=~4\varepsilon ~S ,\tag{H}$$
or at least the Lagrangian 4-form
$$ \delta \mathbb{L}~=~4\varepsilon ~\mathbb{L} .\tag{I}$$
Eq. (H) agrees with OP's eq. (7). The issue is that the Lagrangian density ${\cal L}$ is not a scalar but a density, so there is a non-trivial contribution ${\cal L}d_{\mu}\delta x^{\mu}$ coming from the Jacobian of the integration measure $d^4x$. Therefore the decisive quantity for Noether's theorem is not formula (G) but rather
$$ "\delta {\cal L}"~=~\delta {\cal L}+{\cal L}d_{\mu}\delta x^{\mu} ~=~\delta_0 {\cal L}+  d_{\mu}(\delta x^{\mu}{\cal L})~=~4\varepsilon ~{\cal L} .\tag{J}$$
IV) In conclusion, the infinitesimal transformation (A)-(C) is not a quasisymmetry for the action (D) nor the Lagrangian density (F), and Noether's theorem does not apply.
A: Consider an infinitesimal version of this transformation, given by $\lambda=1+\epsilon$. With this choice, the field and coordinate variations are
$$\delta\phi = \epsilon\phi,\hspace{0.5cm}\delta x=\epsilon x.$$
Under these variations, the action transforms as
$$\delta S=\int_{\Omega'}\mathrm{d}^4x\,\mathcal{L}-\int_{\Omega}\mathrm{d}^4x\,\mathcal{L}.$$
Note that I allowed the same variable $x$ to parametrize both the regions $\Omega$ and $\Omega'$, since these are dummy variables. The only thing that changes is the domain. Thus, the change in the action is
$$\delta S=\int_{\Omega'\setminus\Omega}\mathrm{d}^4x\,\mathcal{L}.$$
If $\Omega$ is a convex set in $\mathbb{R}^4$, then $\Omega'\setminus\Omega$ is a set that is essentially $\partial\Omega$ with thickness $\epsilon$ (this idea can be generalized to non-convex sets, but it becomes a bit more complicated and I really don't want to think about it that much). Thus, in the infinitesimal limit, the integral reduces to
$$\delta S=\epsilon\int_{\partial\Omega}\mathrm{d}^4x\,\mathcal{L},$$
et voilà! Your transformation is now type S2!
I hope this helps!
