If we put a body in a given fluid the buoyant force equalise the
weight of the body. If we increase weight, the buoyant force increase?
Correct. As long as the density of the object is less than or equal to the density of the liquid.
So it must be self adjusting force, is it have any maximum value just
like friction as??
Not exactly sure what you mean by "self adjusting", but the buoyant force does have to increase in order for a heavier object to float. But there is a maximum buoyant force which occurs when the density of the object equals the density of the liquid. I don't see this as analogous to a maximum friction force, by which I assume you mean static friction force. The static friction force does match the applied force to prevent relative motion between the surfaces up to a maximum possible static friction force. Clearly, the buoyant force does not prevent relative motion between the surface of the object and the liquid because the heavier object will submerge further than the lighter object until equilibrium is established.
In any case, in order for an object to float, the upward buoyant force must equal the weight of the object. In other words, in order to float the following equation must be satisfied for all cases where the density of the object is less than or equal to the density of the liquid.
$$V_{o}ρ_{o}g=V_{l}ρ_{l}g$$
The left side of the equation is the weight of the object. $V_o$ is the total volume of the object, which is not necessarily the submerged volume, and $ρ_{o}$ is the density of the object.
The right side of the equation is the buoyant force, which is the weight of the volume of liquid $V_l$ that is displaced by the object times the density of the liquid, $ρ_{l}$.
The greater the weight of the object on the left side of the equation the greater the buoyant force has to be on the right in order for the object to float. The maximum buoyant force occurs when the density of the object equals the density of the liquid, $ρ_{o}=ρ_{l}$, meaning the volume of water displaced equals the volume of the object, $V_{l}=V_{o}$ and the object floats completely submerged. Any further increase in weight results in the object sinking and no further displacement of water, so the buoyant force remains constant. Adding weight simply causes the object to sink faster (assuming the density of the liquid is constant).
Hope this helps.