Does the JVM create a mutex for every object in order to implement the 'synchronized' keyword? If not, how?

Speaking as someone who has looked at the way that some JVMs implement locks ...

The normal approach is to start out with a couple of reserved bits in the object's header word. If the object is never locked, or if it is locked but there is no contention it stays that way. If and when contention occurs on a locked object, the JVM inflates the lock into a full-blown mutex data structure, and it stays that way for the lifetime of the object.

EDIT - I just noticed that the OP was talking about OS-supported mutexes. In the examples that I've looked at, the uninflated mutexes were implemented directly using CAS instructions and the like, rather than using pthread library functions, etc.

You can never be sure that an object will never be used as a lock (consider reflection). Typically every object has a header with some bits dedicated to the lock. It is possible to implement it such that the header is only added as needed, but that gets a bit complicated and you probably need some header anyway (class (equivalent of "vtbl" and allocation size in C++), hash code and garbage collection).

Here's a wiki page on the implementation of synchronisation in the OpenJDK.

(In my opinion, adding a lock to every object was a mistake.)

This is really an implementation detail of the JVM, and different JVMs may implement it differently. However, it is definitely not something that can be optimized at compile time, since Java links at runtime, and this it is possible for previously unknown code to get a hold of an object created in older code and start synchronizing on it.

Note that in Java lingo, the synchronization primitive is called "monitor" rather than mutex, and it is supported by special bytecode operations. There's a rather detailed explanation here.