Is there something similar to Noether's theorem for discrete symmetries?

For continuous global symmetries, Noether theorem gives you a locally conserved charge density (and an associated current), whose integral over all of space is conserved (i.e. time independent).

For global discrete symmetries, you have to distinguish between the cases where the conserved charge is continuous or discrete. For infinite symmetries like lattice translations the conserved quantity is continuous, albeit a periodic one. So in such case momentum is conserved modulo vectors in the reciprocal lattice. The conservation is local just as in the case of continuous symmetries.

In the case of finite group of symmetries the conserved quantity is itself discrete. You then don't have local conservation laws because the conserved quantity cannot vary continuously in space. Nevertheless, for such symmetries you still have a conserved charge which gives constraints (selection rules) on allowed processes. For example, for parity invariant theories you can give each state of a particle a "parity charge" which is simply a sign, and the total charge has to be conserved for any process, otherwise the amplitude for it is zero.

Put into one sentence, Noether's first Theorem states that a continuous, global, off-shell symmetry of an action $S$ implies a local on-shell conservation law. By the words on-shell and off-shell are meant whether Euler-Lagrange equations of motion are satisfied or not.

Now the question asks if continuous can be replace by discrete?

It should immediately be stressed that Noether Theorem is a machine that for each input in form of an appropriate symmetry produces an output in form of a conservation law. To claim that a Noether Theorem is behind, it is not enough to just list a couple of pairs (symmetry, conservation law).

Now, where could a discrete version of Noether's Theorem live? A good bet is in a discrete lattice world, if one uses finite differences instead of differentiation. Let us investigate the situation.

Our intuitive idea is that finite symmetries, e.g., time reversal symmetry, etc, can not be used in a Noether Theorem in a lattice world because they don't work in a continuous world. Instead we pin our hopes to that discrete infinite symmetries that become continuous symmetries when the lattice spacings go to zero, can be used.

Imagine for simplicity a 1D point particle that can only be at discrete positions $q_t\in\mathbb{Z}a$ on a 1D lattice $\mathbb{Z}a$ with lattice spacing $a$, and that time $t\in\mathbb{Z}$ is discrete as well. (This was, e.g., studied in J.C. Baez and J.M. Gilliam, Lett. Math. Phys. 31 (1994) 205; hat tip: Edward.) The velocity is the finite difference

$$v_{t+\frac{1}{2}}:=q_{t+1}-q_t\in\mathbb{Z}a,$$

and is discrete as well. The action $S$ is

$$S[q]=\sum_t L_t$$

with Lagrangian $L_t$ on the form

$$L_t=L_t(q_t,v_{t+\frac{1}{2}}).$$

Define momentum $p_{t+\frac{1}{2}}$ as

$$ p_{t+\frac{1}{2}} := \frac{\partial L_t}{\partial v_{t+\frac{1}{2}}}. $$

Naively, the action $S$ should be extremized wrt. neighboring virtual discrete paths $q:\mathbb{Z} \to\mathbb{Z}a$ to find the equation of motion. However, it does not seem feasible to extract a discrete Euler-Lagrange equation in this way, basically because it is not enough to Taylor expand to the first order in the variation $\Delta q$ when the variation $\Delta q\in\mathbb{Z}a$ is not infinitesimal. At this point, we throw our hands in the air, and declare that the virtual path $q+\Delta q$ (as opposed to the stationary path $q$) does not have to lie in the lattice, but that it is free to take continuous values in $\mathbb{R}$. We can now perform an infinitesimal variation without worrying about higher order contributions,

$$0 =\delta S := S[q+\delta q] - S[q] = \sum_t \left[\frac{\partial L_t}{\partial q_t} \delta q_t + p_{t+\frac{1}{2}}\delta v_{t+\frac{1}{2}} \right] $$ $$ =\sum_t \left[\frac{\partial L_t}{\partial q_t} \delta q_{t} + p_{t+\frac{1}{2}}(\delta q_{t+1}- \delta q_t)\right] $$ $$=\sum_t \left[\frac{\partial L_t}{\partial q_t} - p_{t+\frac{1}{2}} + p_{t-\frac{1}{2}}\right]\delta q_t + \sum_t \left[p_{t+\frac{1}{2}}\delta q_{t+1}-p_{t-\frac{1}{2}}\delta q_t \right].$$

Note that the last sum is telescopic. This implies (with suitable boundary conditions) the discrete Euler-Lagrange equation

$$\frac{\partial L_t}{\partial q_t} = p_{t+\frac{1}{2}}-p_{t-\frac{1}{2}}.$$

This is the evolution equation. At this point it is not clear whether a solution for $q:\mathbb{Z}\to\mathbb{R}$ will remain on the lattice $\mathbb{Z}a$ if we specify two initial values on the lattice. We shall from now on restrict our considerations to such systems for consistency.

As an example, one may imagine that $q_t$ is a cyclic variable, i.e., that $L_t$ does not depend on $q_t$. We therefore have a discrete global translation symmetry $\Delta q_t=a$. The Noether current is the momentum $p_{t+\frac{1}{2}}$, and the Noether conservation law is that momentum $p_{t+\frac{1}{2}}$ is conserved. This is certainly a nice observation. But this does not necessarily mean that a Noether Theorem is behind.

Imagine that the enemy has given us a global vertical symmetry $\Delta q_t = Y(q_t)\in\mathbb{Z}a$, where $Y$ is an arbitrary function. (The words vertical and horizontal refer to translation in the $q$ direction and the $t$ direction, respectively. We will for simplicity not discuss symmetries with horizontal components.) The obvious candidate for the bare Noether current is

$$j_t = p_{t-\frac{1}{2}}Y(q_t).$$

But it is unlikely that we would be able to prove that $j_t$ is conserved merely from the symmetry $0=S[q+\Delta q] - S[q]$, which would now unavoidably involve higher order contributions. So while we stop short of declaring a no-go theorem, it certainly does not look promising.

Perhaps, we would be more successful if we only discretize time, and leave the coordinate space continuous? I might return with an update about this in the future.

An example from the continuous world that may be good to keep in mind: Consider a simple gravity pendulum with Lagrangian

$$L(\varphi,\dot{\varphi}) = \frac{m}{2}\ell^2 \dot{\varphi}^2 + mg\ell\cos(\varphi).$$

It has a global discrete periodic symmetry $\varphi\to\varphi+2\pi$, but the (angular) momentum $p_{\varphi}:=\frac{\partial L}{\partial\dot{\varphi}}= m\ell^2\dot{\varphi}$ is not conserved if $g\neq 0$.

You mentioned crystal symmetries. Crystals have a discrete translation invariance: It is not invariant under an infinitesimal translation, but invariant under translation by a lattice vector. The result of this is conservation of momentum up to a reciprocal lattice vector.

There is an additional result: Suppose the Hamiltonian itself is time independent, and suppose the symmetry is related to an operator $\hat S$. An example would be the parity operator $\hat P|x\rangle = |-x\rangle$. If this operator is a symmetry, then $[H,P] = 0$. But since the commutator of an operator with the Hamiltonian also gives you the derivative, you have $\dot P = 0$.