
It would move Down!
By the definition of the Karman line on wikipedia, the lift force and the "centrifugal force" must be equal to the gravitational force and, therefore, each other. This gives the following equation:
$\frac{1}{2}\rho v^2C_LS = \frac{v^2m}{R_e+h}$
Where $\rho$ is density, v is velocity, $C_L$ is lift coefficient, S is wing area, m is vehicle mass, $R_e$ is the Earth's radius, and h is the altitude.
The velocity terms cancel and I assumed that $C_LS$ and m canceled because S is entirely a design choice and could be chosen such that $C_LS$ was equal to m; regardless, the Karman line should be independent of vehicle (for practical purposes anyway). Following these simplifications, the equation becomes:
$(R_e+h)\rho=2$
Applying the Scale Height model you provided gives a means to compute $\rho$ as a function of altitude. I tried to solve the resulting equation analytically, but It got rather nasty and so I used a zero finder.
Per the "scale height approximation", the density as a function of altitude is the following:
$\rho(h) = \frac{M_{air}P_oe^{-h/H}}{RT_{avg}}$
with $M_{air}$ being the mean molecular mass of the atmosphere, $P_o$ being the s.l. pressure, R being the gas constant, and $T_{avg}$ the mean atmospheric temperature. Pressure ($P_o$) as a function of depth in a fluid scales linearly with the gravitational acceleration- I am assuming the height of the atmosphere to remain the same and the surface pressure to scale with the increase in surface gravity.
$H$ is the scale height and represents the height in which the pressure decreases by a factor of e. It is defined as the following:
$H = \frac{RT_{avg}}{M_{air}g}$
The final equation I used to zero find for the Karman altitude is this:
$(R_e+h)\frac{M_{air}P_oe^{-h/H}}{RT_{avg}}-2=0$ With $P_o$ and $H$ being dependent on surface gravity as defined above.
I first adjusted the "mean atmospheric temperature" until the height was equal to the traditional 100 km mark (with the standard 250K, the value was ~110 km). At 221.55K, I computed a Karman altitude of 100km +/- 5 cm.
Finally I could adjust the Earth's mass. A mass of 1.1 Earth's decreased the Scale Height from 6.47km to 5.88km (effectively increasing the pressure gradient of the atmosphere and providing a lower density at a given altitude) and provided a new Karman Line at 91.5km
EDIT: I would like to say that my algebra in regards to the canceling of $C_LS$ with m is a bit iffy. The Karman line is very dependent on vehicle mass and lift characterisics. For the purposes of simplicity, I was assuming the two values were equal, but realized a moment ago that their units do not cancel. The math still holds, but the algebra should include a unit factor containing the missing dimensions.