One way to look at versions of this problem that involve the host “randomly” opening doors, none of which happen to contain the car, is that this is implicitly eliminating probability mass by discarding cases where one of the randomly opened doors does contain the car.
In the 3 door case, if you pick door 1, there are three possible locations for the car (C1, C2, C3). The host then randomly opens one of the two remaining doors (H2, H3). Each of the six possible combinations is equally likely, but C2+H2 and C3+H3 are implicitly eliminated because in those combinations, the car is revealed and the game ends.
The remaining four cases are C1+H2, C1+H3, C2+H3, and C3+H2. Obviously in this case your odds are 50⁄50 regardless. More significantly, your overall odds remain 1⁄3, because of the eliminated cases. The same logic applies to cases with more doors, where all combinations of C1+H? remain under consideration, while only one combination involving each other car position remains in play.
This differs from the classic statement of the problem because when the host never opens a door with the car, he does not eliminate any of the probability mass for “you picked the wrong door”, thus concentrating the likelihood of “you picked the wrong door” onto a single door, which gives the usual counterintuitive result.
But yes, if you know that the host always opens a door and offers to let the player switch, switching will never make you do worse than not switching. If you don’t know this, there is insufficient information to decide on a strategy.