If we take "charge density" to be the total charge density, including so-called free and bound charges so that $\nabla\cdot{E}=\rho$ (I'm doing this to avoid the mathematical problems with the $D$ field as frequency goes to zero in a good conductor; there's more than one way to handle this), and if we're dealing with a linear, homogeneous, isotropic medium in steady state so that $J=\sigma{E}$ for some constant scalar conductivity $\sigma$, then we get $\rho=\nabla\cdot{J}/\sigma$. But we're at steady state, so the time derivative of $\rho$ is 0, and therefore the divergence of the current density $J$ is also zero because charge is conserved.
Putting this together, this means that the volumetric charge density $\rho$ is zero everywhere inside the wire.
However, you can still get a surface charge density. And in fact you will, because each short piece of the wire (short enough that we can pretend that it has constant voltage) has some capacitance per unit length with respect to an imaginary electrode at infinity. So you'll get a surface charge density at each point that's proportional to the voltage.
It's a little more complicated than that, since each bit of wire is feeling the field not only from the battery but also from all of the surface charges on the other bits of wire. You end up having to solve a partial differential equation to figure out where all the charge has to go in order to set up the steady-state electric field inside the wire. For a kind of normal-sized length of wire that you could hold in your hand, this will take a fair number of nanoseconds, and the amount of charge will be small (like a few picocoulombs per volt). After that, all the time derivatives will be zero, and all the charges will be on the surface of the wire, and the electric field lines will go nicely down the length of the wire, parallel to the current density at every single point, even if the wire diameter changes.
The charge that's associated with this is usually so small that people act like it's zero. Well, it's not quite zero, and in general hooking a wire up to a battery does put a small net charge on the wire. In cases where this actually matters this phenomenon is called "parasitic capacitance." If a circuit needs to operate at very high frequency, this can be a serious problem, because it means that electrical signals can effectively jump from one wire to another. At high enough frequency, a capacitor acts like a short circuit. This is called "capacitive crosstalk."
If you have wire made of two different materials, or if the material is anisotropic, it gets messier. But generally people make wires out of nice, crystallographically symmetric, uniform, metallic materials, and the approximations I'm using should be quite good.
Best Answer
i) Nowhere, as non-zero divergence of current density ($\nabla \cdot \mathbf j$) would mean charge density is changing in time, which would contradict the assumption of stationary flow.
ii) Non-zero divergence of electric field ($\nabla \cdot \mathbf E$) means non-zero density of electric charge (does not need to be point-like though). In metal, the only place charge can be in static situation or during stationary flow assumed is on the surface of the metal object. In fact charge density must be distribution that is not everywhere zero, as the excess charge contributed with its Coulomb field to maintain the electric field and current inside.
The charges on the surface are from the metal itself, and also from those that came from the other parts of the circuit. Their origin does not matter, they can get to the surface so some of them will to in the course of establishing dynamic equilibrium in the metal. These charges are bound only in the sense they cannot escape the metal body, but other than that they are free to relocate anywhere in the body.