Does skewness decrease standard deviation ceteris paribus? For a given probability distribution, probability mass must sum to 1, thus by increasing a parameter corresponding to skewness do you shift probability away from the second central moment (variance) to the tails, thereby reducing standard deviation in general?
 A: NO
Let's develop a counterexample. We will use a $\text{Beta}(1, \beta)$ distribution where $\beta$ controls the skewness and variance.
Equations for variance and skewness of a Beta distribution are taken from the Wikipedia article.
beta_var <- function(a, b){

    return(
        (a*b)
        /
        ((a+b)^2*(a+b+1))
    )
}

beta_skew <- function(a, b){

    return(
        (2*(b - a) * sqrt(a + b + 1))
        /
        ((a + b + 2) * sqrt(a*b))
    )
}

a <- 1
b <- 1
beta_var(a, b) # 0.0833333333333333
beta_skew(a, b) # 0

a <- 1
b <- 2
beta_var(a, b) # 0.0555555555555556
beta_skew(a, b) # 0.565685424949238

a <- 1
b <- 1/2
beta_var(a, b) # 0.0888888888888889
beta_skew(a, b) # -0.63887656499994, magnitude 0.63887656499994

Moving from $\text{Beta}(1, 1)$ to $\text{Beta}(1, 2)$ makes the distribution more skewed while decreasing the variance. Moving from $\text{Beta}(1, 1)$ to $\text{Beta}(1, 1/2)$ makes the distribution more skewed (in the other direction) while increasing the variance.
Therefore, adjusting the skewness can both increase and decrease the standard deviation.
