Sorry I misunderstood you then.

Does your variant looks like this?

1. Omega selects C-sim at random from some distribution over TDT full source-codes.
2. Then Omega selects C-sim-sim at random from the same distribution.
3. Then Omega calculates what will happen if it presents the problem to C-sim, but specifying the simulation's full source code as C-sim-sim. Omega determines the probability of C-sim choosing each of the boxes, conditional on it having seen that fixed C-sim-sim.
4. Then Omega fills the box with lowest probability (of being chosen by C-sim) or uses the tie-break rule.
5. Finally Omega presents the real problem to C-act, but specifying the simulation's full source code as C-sim.

What is the best strategy for TDT to play as C-act?

If that is the problem, then consider the following. It still uses the "unlucky number" construction from the set {1, 2, ..., 10}. Each C-act will always choose its unlucky number with lowest probability, so the money is always in C-sim's unlucky number box.

If C-sim has a different unlucky number from C-act then

 Pick C-sim's unlucky number with probability 1 - epsilon
Pick C-act's unlucky number with probability 0
Pick each of the other boxes with probability epsilon / 8

Else

 Pick the common unlucky number with probability 1/10 - epsilon
Pick each other box with probability 1/10 + epsilon / 9

End If

That looks like winning with probability 9/10 x (1 - epsilon) + 1/10 x (1/10 - epsilon) so close to 91%.

Is there a better strategy though?

P.S. We are getting some interesting behaviour here, with slight variations under the conditions for selecting C-sim and calculating its choice probabilities leading to very different best strategies (and different success probabilities such as 10%, 50%, 91% or close to 100%). Quite fascinating.