Supposing the scale is watertight and supposing that the only restoring force on the scale's top plate is that of the calibrated spring:
The scale will be compressed by the fluid, with or without the person present. The scale will read the weight ($-mg$) of the column of water above a horizontal area equal to the cross-sectional area of the scale's movable surface, plus the weight of whatever is resting on the scale, minus the weight of the water displaced by the object resting on the scale.
Supposing the scale is open to water: the scale will not be compressed by the fluid. The scale will read the weight of the object resting on the scale, minus the weight of the fluid displaced by the object.
Supposing we are out of the water and in an accelerated frame, the scale will rill read the weight of the object, period. The weight of the object, if expressed conventionally as a positive number, is the mass of the object multiplied by the acceleration of the object, $F = ma$. In the technically false, but very useful representation that an object stationary in a gravitational field is indeed stationary and not accelerating: $F = m(a-g)$ where $g$ is how fast the object an object in free fall would accelerate away from the frame, a negative number.
You can set $m_1(a-g) = (m_2-m_1)g$, where $m_2$ is the mass of the displaced water, by varying the density of the object and the fluid.
Algebraically reduce the equation:
$m_1a/g = m_2$
let $m_2 = V\rho_2$
let $m_1 = V \rho_1$
$a/g= \rho_2/\rho_1 $
the ratio of the fluid density $\rho_2$ to object density $\rho_1$ which will cause a scale to read the same as an accelerated frame of acceleration $a$ is the ratio of $a$ to $g$. Since we can only have positive densities, this means we can only match the scales for negative (downward) accelerations.
However, this affects only the force on the scale, not internal forces within the object. The swimmer will still feel like she weighs her normal weight, based on the relative tensions, compressions, and orientation of fluids and tissues in her body, especially her inner ear. Whereas the person on the elevator will actually feel (and be) lighter or heavier depending on her acceleration.