Is it reasonable to use the Kolmogorov-Smirnov test to assess the normality of a random variable? Context
I am confused by the following post where the accepted answer states that :

You can't really even compare the two since the Kolmogorov-Smirnov is
for a completely specified distribution (so if you're testing
normality, you must specify the mean and variance; they can't be
estimated from the data*), while the Shapiro-Wilk is for normality,
with unspecified mean and variance.

*

*you also can't standardize by using estimated parameters and test for standard normal; that's actually the same thing.


Question
Imagine that I have a random sample of measurements X which I standardise using its sample  mean and variance. May I use the Kolmogorov Smirnov test as a GOF test to assess normality of this random sample ?
$$ H_0 : X_{scaled} \sim N(0,1) $$
Illustration
To illustrate my question here is a code snippet in R :
# We wish to do a Goodness of Fit test that X is a random sample from a Normal Distribution N(mu,sigma^2)
X <- c(10.212, 10.103, 10.242, 10.106, 10.102, 10.095, 10.042, 10.093, 10.302, 10.111)
sample.mean <- mean(X)
sample.variance <- var(X)
# Or that standardized X (scaled.X) is a random sample from a standard normal distribution N(0,1)
scaled.X <- (X-sample.mean)/(sqrt(sample.variance))


# Kolmogorov-Smirnov Test H0 : X ~ N(0,1)

ks.test(scaled.X,alternative="two.sided",y = "pnorm")
# Do not reject the null.
# Shapiro Test 
shapiro.test(scaled.X)
# Do reject the null.

Note that the KS test and the Shapiro-Wilk test gives contradictory results, hinting towards the Shapiro Wilk test being more powerful in this specific case. This is however not my main question although any comments on this is gladly welcomed.
The specific area of interest of this question is if using the KS test on a standardized random sample (with  sample statistics) a sound way to evaluate the normality assumption.

Consider the following simulation, in which
the K-S test is asked whether a sample of size
$n=20$ from $\mathsf{Beta}(2,2)$ is normal.
Bogus variants with sample mean and variance
and (similarly) standardizing with sample mean and variance
are also tried. [Approximate variants of the K-S test using sample mean and SD are available.]
Finally, the Shapiro-Wilk test
is asked whether the beta data fit any normal
distribution. (Poor power with small $n.)$
set.seed(2022)
m = 10^5; pv1 = pv2 = pv3 = pv4 = numeric(m)
for (i in 1:m) {
 x = rbeta(50, 2,2)
 pv1[i] = ks.test(x, pbeta, 2,2)$p.val
 pv2[i] = ks.test(x, pnorm, mean(x), sd(x))$p.val
 z = (x -mean(x))/sd(x)
 pv3[i] = ks.test(z, pnorm)$p.val
 pv4[i] = shapiro.test(x)$p.val
}
mean(pv1 <=.05) # aprx K-S P-val for BETA(2,2)
[1] 0.05052       # power aprx same as sig level
mean(pv2 <=.05) # dishonest K-S, for NORM(.5,.2236)
[1] 0.00012       # power below sig. level
mean(pv3 <=.05) # dishonest K-S, for standardized bets
[1] 0.00012       # no power, same as above
mean(pv4 <= 10) # honest Shapiro P-val
[1] 0.15131       # aprx power 15%

The two bogus attempts to trick the K-S test (pv2 and pv3) are
dysfunctional, giving P-values below the nominal significance level of 5%. [As @whuber remarks, the 'standardization' makes the sample look 'too much' like standard normal.]
About 15% of the time, the Shapiro-Wilk test correctly recognizes that
the beta data [all in $(0,1)]$ are inconsistent with any
normal distribution.
Reference: The idea to use $\mathsf{Beta}(2,2)$ in my simulation comes from Figure 1(a) of this relevant paper.
