If you are within the same modifier interval, then it is actually extremely likely that you will get the same stake kernel on both chains. Why wouldn't you?
The modifier interval is a method applied by some PoS coins in order to protect against N@S attacks. Naive PoS schemes in which the next block's hash target is derived from the previous block, are susceptible for this kind of attack. At least this is how I understand Vitalik's blog post:
The issue is this: suppose that you have 1% stake, and thus every block there is a 1% chance that you will be able to produce (hereinafter, sign) it. Now, suppose there is a fork between chain A and chain B, with chain A being the correct chain. The honest strategy is to try to generate blocks just on A, getting an expected 0.01 A-coins per block. An alternative strategy, however, is to try to generate blocks on both A and B, and if you find a block on both at the same time then discarding B. The payout per block is one A-coin if you get lucky on A (0.99% chance), one B-coin if you get lucky on B (0.99% chance) and one A-coin, but no B-coins, if you get lucky on both; hence, the expected payout is 0.01 A-coins plus 0.0099 B-coins if you double-vote. If the stakeholders that need to sign a particular block are decided in advance, however (ie. specifically, decided before a fork starts), then there is no possibility of having the opportunity to vote on A but not B; you either have the opportunity on both or neither. Hence, the dishonest strategy simply collapses into being the same thing as the honest strategy.
I guess what I am saying, is that if there are two chains, why would you not publish the same stake to both chains? If the split happened before the modifier interval has changed (which is the only thing that would alter your utxo hashed kernel value from one chain to the other) then you should have almost identical ability to publish to both chains, assuming that difficulty is relatively similar between the two. You can publish to both chains without modifying a client either, simply all you would need to do is run two nodes, one on each chain.