Warning: this is a long and boring derivation. If you are interested only in the result skip to the very last sentence.
Noether's theorem can be formulated in many ways. For the purposes of your question we can comfortably use the special relativistic Lagrangian formulation of a scalar field. So, suppose we are given an action
$$S[\phi] = \int {\mathcal L}(\phi(x), \partial_{\mu} \phi(x), \dots) {\rm d}^4x.$$
Now suppose the action is invariant under some infinitesimal transformation $m: x^{\mu} \mapsto x^{\mu} + \delta x^{\mu} = x^{\mu} + \epsilon a^{\mu}$ (we won't consider any explicit transformation of the fields themselves). Then we get a conserved current
$$J^{\mu} = {\partial {\mathcal L} \over \partial \phi_{,\mu}} \phi^{,\nu} a_{\nu} - {\mathcal L} a^{\mu} = \left ({\partial {\mathcal L} \over \partial \phi_{,\mu}} \phi^{,\nu} - {\mathcal L} g^{\mu \nu} \right) a_{\nu} .$$
We obtain a conserved charge from it by letting $Q \equiv \int J^0 {\rm d}^3x$ since from $\partial_{\mu}J^{\mu} =0$
we have that
$$ {\partial Q \over \partial t} = \int {\rm Div}{\mathbf J}\, {\rm d}^3 x = 0$$
which holds any time the currents decay sufficiently quickly.
If the transformation is given by translation $m_{\nu} \leftrightarrow \delta x^{\mu} = \epsilon \delta^{\mu}_{\nu}$ we get four conserved currents
$$J^{\mu \nu} = {\partial {\mathcal L} \over \partial \phi_{\mu}} \phi^{\nu} - {\mathcal L} g^{\mu \nu} .$$
This object is more commonly known as stress energy tensor $T^{\mu \nu}$ and the associated conserved currents are known as momenta $p^{\nu}$.
Also, in general the conserved current is simply given by $J^{\mu} = T^{\mu \nu} a_{\nu}$.
For a Lorentz transformation we have
$$m_{\sigma \tau} \leftrightarrow \delta x^{\mu} =
\epsilon \left(g^{\mu \sigma} x^{\tau} - g^{\mu \tau} x^{\sigma} \right)$$ (notice that this is antisymmetric and so there are just 6 independent parameters of the transformation) and so the conserved currents are the angular momentum currents
$$M^{\sigma \tau \mu} = x^{\tau}T^{\mu \sigma} - x^{\sigma}T^{\mu \tau}.$$
Finally, we obtain the conserved angular momentum as
$$M^{\sigma \tau} = \int \left(x^{\tau}T^{0 \sigma} - x^{\sigma}T^{0 \tau} \right) {\rm d}^3 x . $$
Note that for particles we can proceed a little further since their associated momenta and angular momenta are not given by an integral. Therefore we have simply that $p^{\mu} = T^{\mu 0}$ and $M^{\mu \nu} = x^{\mu} p^{\nu} - x^{\nu} p^{\mu}$. The rotation part of this (written in the form of the usual pseudovector) is
$${\mathbf L}_i = {1 \over 2}\epsilon_{ijk} M^{jk} = ({\mathbf x} \times {\mathbf p})_i$$
while for the boost part we get
$$M^{0 i} = \left(t {\mathbf p} - {\mathbf x} E \right)^i $$
which is nothing else than the center of mass at $t=0$ (we are free to choose $t$ since the quantity is conserved) multiplied by $\gamma$ since we have the relations $E = \gamma m$, ${\mathbf p} = \gamma m {\mathbf v}$. Note the similarity to the ${\mathbf E}$, $\mathbf B$ decomposition of the electromagnetic field tensor $F^{\mu \nu}$.
Conservation of baryon number <-> Global gauge invariance
Conservation of lepton number <-> U(1) symmetry
Conservation of strangeness is only for the strong (SU(3) symmetry) and electromagnetic interactions ( local U(1) gauge invariance)
Best Answer
Nature doesn't have this symmetry because your conservation law doesn't hold, either. According to the law of inertia, object keeps on moving with a constant velocity – which is however generically nonzero. In its own rest frame, it's zero, but in other frames, the velocity is nonzero.
If one studies the motion of the center-of-mass, it is indeed moving with a constant velocity. So the conserved quantity that is closest to your "conserved position" is the conserved velocity of the center-of-mass. This conservation law is directly linked, via Noether's theorem, to the Lorentz symmetry of the laws of physics – or, in the non-relativistic limit, to the Galilean symmetry. In the non-relativistic case, the generator of the Galilean symmetry is $\vec x_{\rm cm}$, the center-of-mass position, indeed: the generator of the symmetry is the conserved quantity itself.
If you designed boring laws in which the position has to be conserved, the symmetry would be generated by the conserved quantity $\vec x$. This symmetry generator generates translations in the momentum space. So the laws of physics (the Hamiltonian) would have to be effectively independent of the momentum. That would be pretty bad: you couldn't include the kinetic energy term to the total energy, among other things. That's related to the fact that the particles would have "infinite inertial mass", which would force them to sit at a single point. The whole term "dynamics" would be a kind of oxymoron because things wouldn't be changing with time.
Appendix
Consider the generator equal to the center-of-mass position $$ \vec x_{\rm cm} = \frac{m_1 \vec x_1 + m_2 \vec x_2 +\dots +m_N \vec x_N}{m_1+m_2+\dots +m_N} $$ How do physical observables transform under the symmetry generated by it? Compute the commutators. The commutators of the position above with positions $x_i$ vanish, so positions (at $t=0$) don't transform. However, the commutator with $p_i$ is equal to $m_i \delta_{mn} / M_{\rm total}$, and if this is added to $p_i$ with an infinitesimal coefficient $\vec \epsilon \cdot M_{\rm total}$, you see that all velocities are changed by $$ \vec v_i \to \vec v_i + \vec \epsilon $$ But if all velocities are just shifted by a constant, that's the Galilean transformation. let me emphasize that this simple transformation rule only holds at $t=0$. For $t\neq 0$, one would have to add extra terms proportional to $t$ to the generator (they would be similar for a Lorentz symmetry, too), namely $t\cdot \vec P_{\rm total}$. At any rate, the center-of-mass position is the generator of the Galilean transformations, the transformations switching from one inertial system to a nearby inertial system (which moves by a speed differing by $\delta \vec v$).
Note that the commutator of $\vec x_{\rm cm}$ with the Hamiltonian isn't quite zero, so according to some definitions, it isn't a symmetry. Instead, the commutator is proportional to the total momentum $\vec p$ which is a symmetry itself. So the commutators of various generators yield other generators – the standard form of a Lie algebra (Galilean/Lorentz in this case) in which the Hamiltonian isn't necessarily commuting with everyone else but is one of the generators of a non-Abelian group.