Thanks to Birkhoff's theorem, we know that the field outside a spherically symmetric isolated mass will always be the Schwarzschild metric. In other words, the metric will look like this just outside the surface of the object
$$d s^2 = -\left(1 - \frac{2 M}{r}\right) d t^2 + \frac{1}{1 - 2M/r} dr^2 + r^2 d \Omega^2$$
where $d\Omega^2 = d\vartheta^2 + \sin^2\vartheta d \varphi^2$ is the surface element of a unit sphere. So in some sense the geometry looks the same outside spherical objects in relativity, one only changes the position of the surface and the value of $M$.
However, you will find it surprising that even when you take the same number of particles (say protons and electrons) of a fixed total rest mass $M_0$ and squeeze it into a body of different radius, the resulting value of the parameter $M$ in the metric can be somewhat different.
For instance, when we are talking about a body made out of a perfect fluid, we can use the analysis of Tolmann, Oppenheimer and Volkoff to see that the gravitating mass can be understood to consist of three contributions
$$M = M_0 + \delta M_\mathrm{thermo} + \delta M_\mathrm{binding}$$
$\delta M_\mathrm{thermo}$ corresponds to the internal thermodynamical energy of the gas, and $\delta M_\mathrm{binding}$ is the gravitational binding energy. When we are close to a Newtonian regime, the binding energy can be expressed simply as the Newtonian binding energy (divided by $c^2$)
$$\delta M_\mathrm{binding} = -\int_0^\mathrm{surf.} \frac{G m_0(r) \rho_0 }{r c^2} 4 \pi r^2 dr$$
where $\rho_0$ is the rest-mass density, and $m_0(r) = \int_0^r \rho_0(r') 4\pi r'^2dr'$ is the rest mass contained in the sphere of radius $r$.
