$\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$
Let $\M,\N$ be smooth manifolds, $\M$ oriented and equipped with a Riemannian metric.
Let $L : J^1(\M,\N) \to \mathbb R$ be smooth; We define a functional $E_{\M,\N}:C^{\infty}(\M,\N) \to \mathbb{R}$ as follows:
Given a $C^{\infty}$ map $u : \M \to \N$, we define its prolongation $j^1u : \M \to J^1(\M,\N)$ by $$j^1u(x) = (x, u(x), Du(x)),$$ and define $E$ by the formula
$$ E_{\M,\N}(u)=\int_{\M} L\big(j^1u)\text{Vol}_{\M}.$$
($\text{Vol}_{\M}$ is the Riemannian volume form of $\M$).
Definition: We say $f \in \text{Diff}(\M)$ is a (source) symmetry of $E_{\M,\N}$, if $$ E_{\M,\N}(u)=E_{\M,\N}(u \circ f) \, \text{ for every } \,u \in C^{\infty}(\M,\N)$$
Question: Is it true that $f$ is a local source symmetry? i.e does the equality
$$ E_{f(U),\N}(u)=E_{U,\N}(u \circ f) \, \text{ for every } \,u \in C^{\infty}(f(U),\N)$$
hold for every open subset $U \subseteq \M$?
($U,f(U)$ are subsets of $\M$ so $E_{U,\N},E_{f(U),\N}$ are defined by restricting the domain of integration).
Equivalently, is $f$ a "pointwise-symmetry"- does $$ \big(L(u) \circ f \big)\cdot f^*\text{Vol}_{\M}=L(u \circ f)\cdot \text{Vol}_{\M} \, \,?$$
Edit:
The answer is clearly no without any more restrictions on $L$;
If we take $L\equiv1$, we get $E_{U,\N}(u)=\text{Vol}(U)$ for every $u \in C^{\infty}(f(U),\N)$. This implies any diffeomorphism $f:\M \to \M$ is a source symmetry, but it will be a local symmetry only if $\text{Vol}(U)=\text{Vol}(f(U))$ for every $U \subseteq \M$, i.e only $f$ is volume-preserving.
So, we should assume $L$ is nowhere locally constant (w.r.t the standard topology on the jet space). Is this enough?