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.
You have all the elements in your question, your difficulty is about what is meant by "time reversal symmetry". Time reversal symmetry holds if, when "playing backwards", the motion observed obeys the same law. With friction it is not the case : friction opposes movement, when playing backwards it (seemingly) promotes it.
You can also go to equations for this. Let's have a damped oscillator of mass $m$:
$$
m \frac{\mathrm d^2x}{\mathrm dt^2} = - c \frac{\mathrm dx}{\mathrm dt} - k x
$$
Now play backwards with $\tau=-t$, thus $\mathrm d\tau/\mathrm dt = -1$:
$$
m \frac{\mathrm d^2x}{\mathrm d\tau^2} = c \frac{\mathrm dx}{\mathrm d\tau} - k x
$$
So, the equation is not the same—unless $c= 0$ : the physics of the backward time phenomenon is not the same if there's friction—while it is the same when there is no friction.
Best Answer
Conservation of energy follows from invariance under translation in time, not inversion. This symmetry states that no matter when you do your experiment, it will give the same results. All isolated systems obey this symmetry (and therefore conserve energy) and no violation of it has ever been detected. (Needless to say, it would be a huge event if it were.)
In classical physics, only continuous symmetries - that is, symmetries that can be continuously connected to the identity transformation - have a corresponding conservation law. Quantum physics does permit conservation laws for discrete symmetries but these laws are far harder to visualize.
An example of this is conservation of parity, $P$, which corresponds to invariance under inversion in space, and which gives the parity - even or odd - of wavefunctions. Temporal inversion, $T$, is even harder to turn into a physical quantity because it requires a full relativistic treatment in which time is a coordinate like space and not a parameter (as it is in non-relativistic quantum mechanics). A third discrete symmetry is charge conjugation, $C$, which exchanges particles for their antiparticles.
It turns out that any consistent field theory must be invariant under all three operations when taken together - i.e. under $CPT$. Thus violation of parity - an experiment and its mirror image behaving differently - is possible, for example, if it comes together with violation of $C$ - i.e. the mirror experiment behaves like the original one if it is made of antimatter -, as was discovered in the sixties.
Violations of $C$ and $P$ together have also been discovered in recent years, which means that in some situations violations of $T$ must occur. The recent $B$-meson experiments confirm this. Since the $T$ symmetry does not correspond to energy but to a far more abstract quantity (which is not conserved), this does not lead to a nonconservation of energy.