According to here, nothing bad will happen as long as both USB Type-C ports work to the specification. So option (1) of your list.
To summarise that blog post in case the link ever dies, he essentially explains section 2.3.1 of the Type-C specification:
Image from the linked blog post.
What this says is that unlike earlier USB ports, the USB Type-C specification mandates that power is not applied to the port until CC pin detection is complete. There are basically two pins in the USB Type-C cable which can be used to passively detect whether a device is a host or a slave using resistors without enabling the VBus supply.
Only once a host (DFP) detects that a slave (UFP) has been attached will it start enumeration and enable the VBus supply.
As a result, when you plug two compliant hosts together, nothing at all will happen as neither detects a slave on the other end, so power is not applied (preventing shorting of supplies) and no signalling is performed (preventing bus contention). This would be option (2) from your list.
If however the device is a (cheap) clone of something which is non-compliant to the specification, who knows what will happen. If for example you get a USB Type-C phone charger and it isn't designed to spec, it may always enable the bus voltage, which could cause damage. That is just speculation though.