Return common results in mysql

an agent who has at least two properties in common with two other agents.

Count the number of agents per property (using window functions). Then take the agents where there are at least two properties with two agents:

select agent_id
from (select ap.*,
             count(*) over (partition by property_id) as num_agents
      from agent_properties ap
     ) ap
where num_agents > 2   -- properties that are shared have more than two agents
group by agent_id
having count(*) >= 2;  -- at least two properties that are shared

Here is a db<>fiddle.

EDIT:

Oh, I see. I was going down the wrong path looking at properties first and then agents.

You want a self join with aggregation to calculate the number of properties in common between two agents. Then you want limits in that aggregation:

select agent_id
from (select ap1.agent_id, ap2.agent_id as agent_id2, count(*) as properties_in_common
      from agent_properties ap1 join
           agent_properties ap2
           on ap1.property_id = ap2.property_id and ap1.agent_id <> ap2.agent_id
      group by ap1.agent_id, ap2.agent_id
     ) aa
where properties_in_common >= 2
group by agent_id
having count(*) >= 2