Does a symmetry necessarily leave the action invariant? 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.
 A: Comments to the question (v2):


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*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
$$\tag{8} Q~=~m\dot{q}t-mq. $$
