Well, I thought you were simply referring to the post that you clicked on the "quote" tab, in which I never actually mentioned a short circuit. That was in another post to the OP a day before so confusion resolved.
I still dispute that would blow the cars fuse as the 5v supply is derived from an electronic power supply consisting in basic terms of a primary 12v side connected to the car's fuse and a secondary side 5V connected to in the case of the OP, a dash cam and in my case when my car did nearly catch fire, a mobile phone. As the fault was/would be seen on the 5v portion then the 5v part of the electronics would happily continue pumping out 5V at whatever current it could muster until the electronic components in the 5v section expired as the fault current that would be seen in the 5v section would not be seen to the same magnitude on the 12v side. So using the Ohms Law, lets for the sake of discussion say that the dash cam normally takes 130mA, so thats 0.13A @ 5V = .13 x 5 = 0.65 watts on the output side and assume the whole process is 100% efficient, which it won't be, that equates .65wats / 12v =0.054A being consumed from the 12v battery and assuming the fuse is rated at 5A it is therefore going to take a lot more for that fuse to blow.
Now lets assume that the 5v side suddenly sees an overload of 50 times its normal current flow due to a short circuit close to the mini USB socket on the dashcam, that means that the cable is now carrying a current of 6.5A and using ohms law, Volts times current = watts. So that's 5v x 6.5A = 32.5Watts. So 32.5 watts at 12 v = 32.5 divided by 12 = 2.708A, still way within the 5A rating of the fuse, so the fuse would still not blow.
Now if we factor in the losses because no conversion process is 100% efficient, so let's say that it is only 30% efficient, and we look at the those calculations again, we get the following:-
The healthy circuit's 5v side remains the same so that's 5V, .13A and .65watts and the 12v side equates to 12v, .65/.7 =.928Watts and the current then becomes .928Watts / 12v = 0.077A
Under the 30% efficiency but with 50 times factor of increased current on the 5v side, the 5v side is as before 5v x 6.5A = 32.5Watts, but the 12v side now has to have the losses added to it the 32.5W becomes 46.428watts and the current is 46.428/12=3.869Amps and still within the 5A rating of the fuse so still not going to blow.
Now if we could put a fuse in the 5v circuit, inside the plug along with the 12v to 5v convertor and rate the fuse at 2A then yes that fuse would blow and protect the circuit from a fire risk. Some of those 12v power plugs do have a lower fuse in the centre spring-loaded contact which is on the 12v side, but the cheaper ones don't and by the OPs own admission, the dashcam is only a cheap one, so probably does not so is therefore completely at the mercy of the cars fuse, of which I am pretty sure is a 20A rated one to handle a air compressor being plugged to inflate the tyres will have a pretty big surge current when the motor starts up.