Consider a coin independently tossed 10 times. Assume under the measure $P$, $Pr(H)$ > 0.5 but not equal to 1.
Let a risk neutral person be iteratively given the gamble between getting atleast $n$ heads versus atleast $n$ tails. Clearly the person always chooses getting atleast $n$ heads. Let $X$ represent number of heads and $Y$ the number of tails. Thus we have
$E(1_{X>=n})>E(1_{Y>=n})$ for all $n$, and thus $X$ FOSD the $Y$ under this probability measure.
Now change to a measure where $Pr'(H)< 0.5$ but not equal to 0. Specify event probabilities by retaining the independence assumption. This is equivalent to the previous measure, as all event sets having 0 or 1 probability in the former retain the same probability in the latter.
Now consider the same gamble given to the same person. Now, the signs get flipped! $E'(1_{X>=n})<E'(1_{Y>=n})$ for all $n$, so now the $Y$ FOSD $X$.
So the assertion is false.
When does it preserve FOSD? A look at some special cases below:
If the RV's are such that the minimum of $X$ exceeds the maximum of $Y$, then the stochastic dominance will be maintained under any change of measure, because the $Pr(w: X(w)>X_{min})=0$ and $Pr(w:Y(w)<Y_{max})=1$ under any equivalent measure, so the distributions again are non intersecting.
For discrete RV's $X$ and $Y$, a sufficient condition for violation of FOSD order is for the means to change order. This is because the mean can be written as the sum across all $u$, of $Pr(X>=u)$.
What can the Radon-Nikodym derivative (RND) look like so that FOSD is not violated?
By simple change of measure using the RND $Z$,
$E'(1_{X>=u})=E(1_{X>=u})+cov(Z,1_{X>=u})$
and
$E'(1_{Y>=u})=E(1_{Y>=u})+cov(Z,1_{Y>=u})$
and thus it suffices that: $cov(Z,1_{X>=u})>cov(Z,1_{Y>=u})$ for all $u$.
Intuitively, $Z$ has more of $X$ in it than $Y$, or $X$ is a better predictor of $Z$ than $Y$. It also suffices of Z is independent of both X and Y.
