In the limit where the solenoid is very long, not much actually happens to the field.
In general to solve a problem like this, we exploit the linearity of Maxwell's equations to write the total magnetic field as $\vec{B} = \vec{B}_\text{solenoid} + \vec{B}_\text{induced}$, where $\vec{B}_\text{solenoid}$ is the field solenoidal field that you already know, and the second term is whatever comes from the magnetization of the metal.
When a simple (linear and homogeneous and isotropic) material is placed inside a static magnetic field, the induced field is determined by the relative permeability $\mu_r$ of the material. In an external field $\vec{H}$, the total field becomes $\vec{B} = \mu_0 (\vec{H} + \vec{M}) = \mu \vec{H}$, where $\mu = \mu_0 \mu_r$.
If the metal is para- or diamagnetic, $\mu_r \approx 1$ and not much happens; the metal only has a very weak magnetic response, and its presence doesn't really change anything. So let us turn to ferromagnetic metals.
The magnetic response of a ferromagnetic material is far from linear, and if the solenoidal field is weak, and the metal was already magnetized, the metal might keep its original magnetization independently of the solenoidal field. However, if we assume that the metal was not already magnetized, or that the solenoidal field is strong enough to change the magnetization axis, then we can still model the field in the metal as $\vec{B} = \mu \vec{H}$ for some effective (field-strength dependent) $\mu$.
In this case the induced field will point along the $z$-direction. The standard electromagnetic boundary conditions demand that the tangential component of $\vec{H}$ be continuous across an interface; therefore at the interface between the metal and vacuum, we have the boundary condition
$$\vec{H}_\text{vac} = \vec{H}_\text{metal}, $$
along with the consitutive relations
$$ \vec{B}_\text{vac} = \mu_0 \vec{H}_\text{vac}, $$
$$ \vec{B}_\text{metal} = \mu \vec{H}_\text{metal}. $$
The solution to these equations is simply $H_\text{vac} = \vec{H}_\text{metal} = H_\text{solenoid}$, i.e. the field in the vacuum region is entirely unaffected by the presence of the ferromagnet. Note that this would not be the case if we included the finite size of the system in our analysis; in this case, the boundary condition for the normal component of the $B$ field at the ends of the ferromagnetic would result in an external field. We can understand this conclusion by thinking of the metal in our analysis as a simple ferromagnet with poles that are infinitely far away (located at $z = \pm \infty$), and because of the large distance to the poles, the external field at our analysis point is infinitely weak.
However, the presence of the metal changes the total magnetic energy of the system. The magnetic energy is given by $E = \frac{1}{2} \int_V \vec{H} \cdot \vec{B} \text{d}r$, where the integral is over all of space. Because the $B$-field inside the metal is stronger than in vacuum ($\mu > \mu_0$), the total energy of the system will be larger in the presence of the ferromagnetic metal. This is why inductors are designed as coils wound around a ferromagnetic core - maximizing the inductance is equivalent to maximizing the magnetic energy of the system.
