What's wrong with this Wheatstone bridge?
Below; just another way to see what's been written in comments.







simulate this circuit &ndash; Schematic created using CircuitLab

Your first claim in your question is wrong. The current \$i\$ in the top schematic you provide isn't \$9\:\text{A}\$. Instead, it is \$3\:\text{A}\$ (obviously.) You can easily see this by comparing the currents entering and leaving.

(Swapping \$R_1\$ with \$R_2\$ in the above, rightmost case would arrange things so that the currents flowing into and out of the left side would be balanced. Same with the currents flowing into and out of the right side. So in that case there would be no need for a current between the left and right sides because ... well, because you've changed the question being asked now.)



Now, let's adjust the above, rightmost schematic:







simulate this circuit

Again, without having to introduce any new resistors, you can see the obvious. The \$9\:\text{A}\$ flowing downward from above must obviously also flow along the wire connecting top to bottom. There's no other way for it.

(Swapping \$R_1\$ with \$R_2\$ in this last case makes no difference. The same current sum still must flow through them to ground. So in this case you can swap \$R_1\$ and \$R_2\$, or swap \$R_3\$ and \$R_4\$, without changing the current arriving and leaving through the narrow vertical wire connecting the pairs.)