Field diffeomorphisms preserve the equations of motion In this paper (Field Diffeomorphisms and the Algebraic Structure of Perturbative Expansion, by Kreimer & Velenich), the authors claim in section 3, page 3, that the field diffeomorphism $F(\phi)$ given by:
$$F(\phi) = \sum_{k=0}^\infty a_k \phi^{k+1}\tag{1}$$
"preserve Lagrange's equations" when applied onto a real, massless scalar field with Lagrangian:
$$\mathcal{L}(\phi,\partial_\mu\phi) = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi. \tag{2}$$
This leads to the following transformed Lagrangian:
$$\begin{align} \mathcal{L}_\text{F} (\phi,\partial_\mu\phi) &= \mathcal{L} (F(\phi),\partial_\mu F(\phi)) \\ & = \mathcal{L} (F(\phi),F'(\phi) \partial_\mu \phi) \\ & = \frac{1}{2} F'(\phi)^2 \partial_\mu \phi \partial^\mu \phi. \tag{3} \end{align}$$
The equations of motion of $(2)$ are as usual given by:
$$\partial_\mu \partial^\mu \phi = 0, \tag{4}$$
and I would like to check that this is indeed conserved by the field diffeomorphism given in $(1)$. First I calculated $F'(\phi)$:
$$F'(\phi) = \sum_{k=0}^\infty a_k (k+1) \phi^k, \tag{5}$$
leading to:
$$\mathcal{L}_\text{F} (\phi,\partial_\mu\phi) = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi \sum_{k,l} a_k a_l (k+1)(l+1) \phi^{k+l} \tag{6}$$
The equations of motion are then:
$$\begin{align} \partial_\mu \frac{\partial\mathcal{L}_\text{F}}{\partial(\partial_\mu\phi)} - \frac{\partial\mathcal{L}_\text{F}}{\partial\phi} &= \sum_{k,l} a_k a_l (k+1)(l+1) \left[\partial_\mu \left( \partial^\mu \phi \phi^{k+l} \right) - \frac{1}{2} (\partial \phi)^2 (k+l) \phi^{k+l-1}\right] \\ &= \sum_{k,l} a_k a_l (k+1)(l+1) \phi^{k+l} \left[ \partial_\mu\partial^\mu \phi + (k+l)\frac{1}{\phi} (\partial\phi)^2 \right] \\ &\overset{!}{=} 0 \tag{7}\end{align}$$
This is not the same thing as $(4)$,right? And in a way, I think that it would be quite surprising if it were. Should it be the same equation of motion? And if not, what is meant by "preserving Lagrange's equations"?
 A: What the preservation of Lagrange's equations means in this case, is that the equations of motion for $\phi$ after the diffeomorphism $F$ are automatically satisfied if the equations of motion for $\phi$ before the diffeomorphism are fulfilled. In other words, there is not need to impose
$$\partial_\mu \frac{\partial\mathcal{L}_\text{F}}{\partial(\partial_\mu\phi)} - \frac{\partial\mathcal{L}_\text{F}}{\partial\phi} = 0\tag{8}$$
In the example above, it is easy to show that:
$$\partial_\mu \frac{\partial\mathcal{L}_\text{F}}{\partial(\partial_\mu\phi)} - \frac{\partial\mathcal{L}_\text{F}}{\partial\phi} = F'(\phi) \left[\partial_\mu \frac{\partial\mathcal{L}_\text{F}}{\partial(\partial_\mu F(\phi))} - \frac{\partial\mathcal{L}_\text{F}}{\partial F(\phi)} \right]\tag{9}$$
and this is equal to $0$ since
$$\partial_\mu \frac{\partial\mathcal{L}_\text{F}}{\partial(\partial_\mu F(\phi))} - \frac{\partial\mathcal{L}_\text{F}}{\partial F(\phi)} = \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)} - \frac{\partial\mathcal{L}}{\partial \phi} \overset{!}{=} 0\tag{10}$$
This implies that the diffeomorphism does not generate new physics, at least at tree level.
