A symmetry maps a configuration with stationary action to another configuration with stationary action. However, does it necessarily preserve the value of the action exactly? It seems that it should be possible for a symmetry to map a state with action $S$ onto a state with action $S + k$ where $k$ is a constant, for example; if it does this, then a configuration of stationary action will always be mapped to a configuration of stationary action. However I wasn't able to construct any examples of this type. All examples of physical symmetries that I know of leave the action invariant.
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if we want to apply Noether's theorem, the infinitesimal transformation needs to leave the action invariant (up to a divergence term) – Christoph Feb 15 '15 at 10:00
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Comments to the question (v2):
First of all, recall the notion of an (off-shell) quasi-symmetry. It means that the action $S[\phi]$ changes by a boundary integral under the transformation of the fields $\phi$ and spacetime point $x$, cf. e.g. this and this Phys.SE posts.
Since the action $S=\int_R \!d^nx~{\cal L}$ is an extensive variable, it is clear that OP's constant $k$ in a transformation of the form $$\tag{1}S~\longrightarrow~S^{\prime} ~=~ S+k$$ is also an extensive variable, e.g. it depends on the spacetime region $R$. So the constant $k$ is not just a number. It depends on various parameters. To make sense of OP's proposal, we assume that the constant $k$ does not depend on dynamical active bulk data, but only on boundary data fixed by boundary conditions.
An important question is whether the transformation (1) can be realized on the fields $\phi$ and spacetime point $x$.
If the transformation (1) is realized, we conjecture that it is always as a quasi-symmetry.
Example: Consider a free point particle $$\tag{2}S~=~\int_{t_i}^{t_f} \!dt~L, \qquad L~=~\frac{m}{2}\dot{q}^2,$$ with Dirichlet boundary conditions $$\tag{3} q(t_i)~=~q_i\quad\text{and}\quad q(t_f)~=~q_f. $$ Now consider the transformation $$\tag{4} q(t)~\longrightarrow~ q^{\prime}(t) ~=~ q(t) + \varepsilon t. $$ The transformation (4) is a quasi-symmetry of the Lagrangian $$\tag{5} L~\longrightarrow~ L^{\prime} ~=~L + \frac{dF}{dt}, \qquad F~=~\varepsilon mq +\frac{\varepsilon^2m}{2}t,$$ and a quasi-symmetry of the action $$\tag{6}S~\longrightarrow~S^{\prime} ~=~ S+k,$$ $$\tag{7}k~=~F(t_f)-F(t_i)~=~\varepsilon m(q_f-q_i) +\frac{\varepsilon^2m}{2}(t_f-t_i).$$ The corresponding conserved Noether charge is $$\tag{8} Q~=~m\dot{q}t-mq. $$
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But can we construct such a transformation on the solutions to the EOM? i.e., what Lagrangian and what physical transformation of states would yield that change in the Lagrangian? – Brian Bi Feb 15 '15 at 16:18
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