Holding the density constant and varying the orientation, the height of the GC of the body varies.
We imagine some torque but not lift on the body to prevent motion.
The volume of the displaced water is constant but the depth of the CG of the displaced water also varies.
I don’t know if these two distances are the same.
<p>
Consider a body with two small volumes, one twice the other.
They are held apart by a massless and a volumeless rod.
They are constrained vertically with the large volume down.
When the density is ⅔ any vertical position with the large mass under water and the smaller above is in equilibrium.
This is perhaps a paradoxical configuration.
Skip it for now.
<p>Consider a thin isosceles triangle constrained to float vertically.
Its height is 1 and its volume is ½.
It is constrained to float tip up.
A table of Height of  density = ρ; CG of triangle = h;
moment of displaced water = mw;
depth of CG of displaced water = H.
<table>
<tr><td>ρ<td>h<td>mw<td>H
<tr><td>0<td>⅓<td>0<td>0
<tr><td> ¾ <td>−1/6<td>5/48<td>−5/36
<tr><td>1<td>−⅔<td>⅓<td>−⅔
</table>
This is not the pattern I expected.
<hr>
integral from 0 to ½ of x(½ + x) = I x^2 + ½x = (x^3)/3 + ¼ x^2
= 1/24 + 1/16 = 5/48
<br> m = gc*v
<br> 5/48 = H*(¾); H = 5/48 4/3 = 20/144 = 10/72 = 5/36