Is your position in a distribution determined by the sample size? I'm not sure how to describe this question in mathematical terms concisely. To help state the question let me use an example:
Assume I play a game and I belong to the top 10% of players on my server. Assuming the skill of the server population follows a normal distribution and all servers have an equal distribution (there is not one server with more better players than others).
The question is: Knowing that I'm among the top 10% of players on my server can I state that I'm among the top 10% of all players?
I believe this question is identical to "If I'm in the top percentile of a distribution and I add more samples, will I remain in the top percentile?" - Is this correct?
 A: Roughly the same, but you can't expect your
standing to be exactly the same when the
group size is increased. Also, large normal samples
have more opportunities to generate high (and low)
outliers, so it may be you'll rank a little
lower in the large group.
The following simulation tries this out for
the 95th percentile when your small group of 100
players is merged into a larger group to get 1000
players altogether.
set.seed(2022)
sml = sort(rnorm(100, 50, 7))
sml.95 = sml[95]; sml.95
[1] 60.33687

If you're at the 95th percentile of your group
of 100, then your score would be $60.34.$
Now merge your small group with a larger one of 900
players from the same normal distribution, to make
a large group of 1000 players altogether.
lrg = sort(c(sml, rnorm(900, 50, 7)))
lrg.95 = lrg[950]; lrg.95
[1] 61.17881

So, the player at the 95th percentile of the large
group of 1000 will have a higher score of $61.18 > 
60.34$ than yours and your percentile would go
down by a bit. (Gaining would also be possible.)
mean(lrg <= sml.95)
[1] 0.937

To see if you lose a bit more often than you gain
a bit, we can see what happens over 100,000 such
mergers.
set.seed(116)
m = 10^5;  sgn = numeric(m)
for (i in 1:m) {
 sml = sort(rnorm(100, 50, 7))
 sml.95 = sml[95]; sml.95
 lrg = sort(c(sml, rnorm(900, 50, 7)))
 lrg.95 = lrg[950]; lrg.95
 sgn[i] = sign(lrg.95-sml.95)
}
table(sgn)
sgn
   -1     0     1 
38406  1877 59717 

It seems your percentile would go down a bit
slightly more than half the time.
For full disclosure, I will admit that several
slightly different definitions of quantiles are
in common use. By counting the actual position (95 or 950 out of the sorted sample)
as the 95th percentile, I have tried to use a
'fair' definition for the purpose at hand, but it
is possible that other methods of determining
quantiles might give slightly different results.
