Another way (without Nulls and without cycles in the FOREIGN KEY relationships) is to have a third table to store the "favourite children". In most DBMS, you'll need an additional UNIQUE constraint on TableB.

@Aaron was faster to identify that the naming convention above is rather cumbersome and can lead to errors. It's usually better (and will keep you sane) if you don't have Id columns all over your tables and if the columns (that are joined) have same names in the many tables that appear. So, here's a renaming:

Parent
    ParentID        INT NOT NULL PRIMARY KEY

Child
    ChildID         INT NOT NULL PRIMARY KEY
    ParentID        INT NOT NULL FOREIGN KEY REFERENCES Parent (ParentID)
    UNIQUE (ParentID, ChildID)

FavoriteChild
    ParentID        INT NOT NULL PRIMARY KEY
    ChildID         INT NOT NULL 
    FOREIGN KEY (ParentID, ChildID) 
        REFERENCES Child (ParentID, ChildID)

In SQL-Server (that you are using), you also have the option of the IsFavorite bit column you mention. The unique favourite child per parent can be accomplished via a filtered Unique Index:

Parent
    ParentID        INT NOT NULL PRIMARY KEY

Child
    ChildID         INT NOT NULL PRIMARY KEY
    ParentID        INT NOT NULL FOREIGN KEY REFERENCES Parent (ParentID)
    IsFavorite      BIT NOT NULL

CREATE UNIQUE INDEX is_FavoriteChild
  ON Child (ParentID)
  WHERE IsFavorite = 1 ;

And the main reason that your option 1 is not recommended, at least not in SQL-Server, is that the pattern of circular paths in the foreign key references has some problems.

Read a quite old article: SQL By Design: The Circular Reference

When inserting or deleting rows from the two table, you'll run into the "chicken-and-egg" problem. Which table should I insert first - without violating any constraint?

In order to solve that, you have to define at least one column nullable. (OK, technically you don't have to, you can have all columns as NOT NULL but only in DBMS, like Postgres and Oracle, that have implemented deferrable constraints. See @Erwin's answer in a similar question: Complex foreign key constraint in SQLAlchemy on how this can be done in Postgres). Still, this setup feels like skating on thin ice.

Check also an almost identical question at SO (but for MySQL) In SQL, is it OK for two tables to refer to each other? where my answer is pretty much the same. MySQL has no partial indexes though, so the only viable options are the nullable FK and the extra table solution.

• Option 1

  *This doesn't seem like a great idea to me, because it complicates the SQL to select all properties applied to a feature…

  It does not necessarily complicate query SQL (see conclusion below).

  …and doesn't readily scale to more conditions…

  It scales readily to more conditions, as long as there are still a fixed number of conditions, and there aren't dozens or hundreds.

  However, it does enforce the requirement of a certain number of conditions per (X,Y) pair. In fact, it is the only option here that does so.*

  It does, and although you say in a comment that this is "the least important of my requirements", you haven't said it doesn't matter at all.

• Option 2

  One downside to this is that it doesn't specify the number of conditions for each pair. Another is that when I am only considering the initial relationship…I then have to add a DISTINCT clause to avoid duplicate entries…

  I think you can dismiss this option because of the complications you mention. The objx_objy table is likely to be the driving table for some of your queries (eg "select all properties applied to a feature", which I am taking to mean all properties applied to an objx or objy). You can use a view to pre-apply the DISTINCT so it is not a matter of complicating queries, but that's going to scale very badly performance-wise for very little gain.

• Option 3

  Does it make sense though to create a new ID that identifies nothing other than existing IDs?

  No, it doesn't — Option 4 is better in every regard.

• Option 4

  …it basically duplicates an entire table multiple times (or feels that way, anyway) so also doesn't seem ideal.

  This option is just fine — it is the obvious way of setting up the relations if the number of properties is variable or subject to change

Conclusion

My preference would be option 1 if the number of properties per objx_objy is likely to be stable, and if you can't imagine ever adding more than a handful extra. It is also the only option that enforces the 'number of properties = 3' constraint — enforcing a similar constraint on option 4 would likely involve adding c1_p_id… columns to the xy table anyway*.

If you really don't care much about that condition, and you also have reason to doubt that the number of properties condition is going to be stable then choose option 4.

If you aren't sure which, choose option 1 — it is simpler and that is definitely better if you have the option, as others have said. If you are put off option 1 "…because it complicates the SQL to select all properties applied to a feature…" I suggest the creating a view to provide the same data as the extra table in option 4:

option 1 tables:

create table prop(id integer primary key);
create table objx(id integer primary key);
create table objy(id integer primary key);

create table objx_objy(
  x_id integer references objx
, y_id integer references objy
, c1_p_id integer not null references prop
, c2_p_id integer not null references prop
, c3_p_id integer not null references prop
, primary key (x_id, y_id)
);

insert into prop(id) select generate_series(90,99);
insert into objx(id) select generate_series(10,12);
insert into objy(id) select generate_series(20,22);

insert into objx_objy(x_id,y_id,c1_p_id,c2_p_id,c3_p_id)
select objx.id, objy.id, 90, 91, 90+floor(random()*10)
from objx cross join objy;

view to 'emulate' option 4:

create view objx_objy_prop as
select x_id
     , y_id
     , unnest(array[1,2,3]) c_id
     , unnest(array[c1_p_id,c2_p_id,c3_p_id]) p_id
from objx_objy;

"select all properties applied to a feature":

select distinct p_id from objx_objy_prop where x_id=10 order by p_id;

/*
|p_id|
|---:|
|  90|
|  91|
|  97|
|  98|
*/

dbfiddle here