For starters, here is an artificial example (apart from the trivial one, where you have 0% or 100% equity) where running it twice does not change variance:
You have: A A
Opponent has: 5 4
Board: T 9 8
Dealer has been extremely clumsy and all other cards are dead except: 9 9 8 8

If you get a full house in the first run, then you will get a full house in the secon run (and vice versa), so running it once or twice doesn't make any difference.

Could a similar situation be made where running it once is actually a less variance game than running it several times? If not, how do you prove it mathematically?

I think that the latter is actually the case, but I don't know how to prove it.

If we are free to stack the deck we can manipulate the variance. If we are not allowed to stack the deck, we cannot decrease variance by running it twice.

In this spot theres a 50% chance we win/lose a stack when we run it once. And a 50% chance we win/lose a stack when we run it twice. So its the same.
I cant think of a hypothetical situation where it would be higher.
When we run it twice , a probability of chopping it is introduced, which will be 1 - [ P [hand 1 wins both runs] + P [hand 2 wins both runs]].

When there is no chance of chopping it, running it once and twice become the same, but this is the closest running it twice can become variance wise to running it once.

Note: when I say chopping it I refer to winning one run and losing the other. Im not taking into account stuff like AK vs QQ on a 34567 board

Edit: I can try come up with a better mathematical argument later, on my phone atm

The variance cannot be higher running it twice. It will be the same only in the case where if you win the first time, you must win the second time as in your example.

var(X) = E(X^2) - E^2(X)

Let
f = P(win first run)
s = P(win second run after winning first run)
W = pot size
E = expected value which is the same for running once or twice.

var(run once) = f*W^2 - E^2

var(run twice) = f*s*W^2 + 2*f*(1-s)*(W/2)^2 - E^2

The 2 appears in the second equation since the probability is the same if we win the first and lose the second or lose the first and win the second because the same cards could have come in the other order with the same probability. You could get this same result by summing the variances for the 2 runs (which are the same before both runs) and adding twice the covariance.

var(run twice) >= var(run once) would mean

f*s*W^2 + 2*f*(1-s)*(W/2)^2 >= f*W^2.

This reduces to

s >= 1.

So this can only be met with equality.

Thanks, BruceZ, very nice. Seems the 2 player 2 runs (and no chop) case is solved. Does it generalize to more players (and different chops allowed) and more runs?

I think the more players and chops case can be reduced to one where there are possible k different outcomes (pot sizes) W_1, W_2, ..., W_k each happening with probability p_1, p_2, ..., p_k, respectively.

Yes -- the most extreme possible outcome, if several people run a board several times, is for the same person to win every run. When you only run it once, only those extreme results happen; when you run it more than once or make any kind of a chop, some intermediate (lower variance) results become possible.

I imagine one could make a formula for n-way races, but I doubt the formula would be particularly enlightening.