You are postulating a quite unlikely situation except by intentional mis-use/experimentation of the CAN bus. In general, (2) nodes won't be EXACTLY synchronized and even if they differ in timing by some reasonable fraction of a bit width, one or the other (the first to start with a dominant bit) will win, based on the sample points within the bit widths. If you did intentionally, exactly, synchronize (2) nodes with the exact same message and sent them at exactly the same time, there would in essence, only be (1) message on the bus (the exact same one) even through it came from (2) sources. There are times when we do want to send a broadcast message out for all (or many) nodes on a CAN bus to receive at the same time, but on the same bus, we generally want the messages to be identifiable as to their source - this is commonly used with the producer-consumer model.
In this model, it is common for a producer (message sender) to issue a message that will be simultaneously received and used by multiple consumers. If you have multiple producers which are producing the exact same message at the exact same time, there is no advantage/use to it. The consumer nodes cannot tell the difference. The principle advantage of CAN is that a single message can be received/used by multiple nodes at the same time, contributing to the efficiency of the system.