First off, I wouldn't re-wire anything. You'll immediately void any warranty if you do that. Also you better check your owners manual. If it says the minimum ohm load is 4Ω, you're playing with fire trying to run it at 3.2Ω and liable to end up with an expensive paper weight because all the magic smoke in your amp just escaped (interior components burned up).
Also, you seem to be a bit confused on the way parallel and series connections work.
Series connections add ohms together increasing ohms load - parallel connections reduce ohms load. If you have a 210 cab rated for 8Ω, that cab is using either two 4Ω speakers wired in series, which increases the ohms load (4Ω+4Ω in series=8Ω load) or it's using two 16Ω speakers wired in parallel, which decreases the ohms load (16Ω+16Ω wired in parallel=8Ω load. If the internal speakers are 16Ω speakers, when you wire them internally in series, that will give you a 32Ω cab (16Ω+16Ω in series=32Ω load).
The combination of an 8Ω cab and a 32Ω cab hooked up in parallel (industry standard connection) will give you a 6.4Ω load, which your amp will handle That means that less than half of that 1000-watts into 4Ω load rating will actually go out to the speakers to be split between them. I'm guessing 650 to 700 watts or so? And the 8Ω cab will get 4 times as much power as the 210. So the 210 would not really be adding a lot at that point.
If you really wanted to get the same amount of power to each speaker using a 410 cab and a 210 cab, you could do that by using an 8Ω 210 and a 4Ω 410 cab IF and only IF your amp can handle a 2.67Ω load. Normally an amp will divide the power evenly between two cabs of the same ohm rating. With a 4Ω and 8Ω split, the 4Ω will get twice as much power as explained earlier. For ease of explanation, lets pretend your 1000-watt amp that will handle a 2.67Ω minimum ohms load. It can't, but we'll pretend it does so you can understand how this works
The cab with the lower ohms rating (4Ω instead of 8Ω) will get twice as much power as the 8Ω cab. That means the 4Ω cab will get 666.67-watts and the 8Ω cab will get 333.33-watts. Divide 666.66 by 4 and you get 166.66 watts per speaker in your 410. Divide 333.33 by 2 and you also get 166.66 watts power speaker. That would mean all your speakers would be getting exactly the same amount of power. That's actually pretty cool IF your amp could handle a 2.67Ω load - it can't.
Another caveat would be that your 4Ω 410 cab would have to be able to handle 667 watts (rounded the fraction of a watt up). Had some folks do that with Eden gear when I was a moderator at the Eden forum, but their amps could handle a 2Ω load, so a 2.67Ω load wasn't a problem. Their cabs could handle either 700 or 1000-watts depending upon whether it was a D410XLT or a D410XST. Any of the D-series 210 Eden models could handle 267-watts.
Now, if you have two 8Ω cabs then your power will be split equally between the two - each cab would get 500-watts. As long as your cabs can handle 500-watts, that will work. Your 210 c ab will be getting the same amount of power as the 410 but it will only be split by two speakers, not four. So each of the 210's speakers will be getting twice as much power per speaker as your 410 will be, but if your 210 is rated for 500-watts rms, it will work. In cases like that, it's wise to put the 210 on top so it is easiest for you to hear if the speakers start getting stressed. If you are using distortion in your signal chain, you won't be able to hear the speakers being stressed before they're blown. And, even with a clean signal, if you put the 210 on the bottom you'd probably blow the speakers before you knew there was a problem because your ears are on your head, not down by your ankles.
