Is there a measure of spread/scale which has this property after the distribution is transformed? Suppose we have a continuous random variable $X$ and a transform $f(x)$ which is strictly increasing. Let $Y=f(X)$ be the variable after transforming.
If $m_x$ and $m_y$ are the medians of $X$ and $Y$ then we have a nice property that $m_y=f(m_x)$. This should be obvious because 50% of the values of $X$ are below $m_x$ and for all $x<m_x$ we have $f(x)<f(m_x)$.
Is there any measure of spread/scale which has this property? i.e. For the spread $s$ we have $s_y=f(s_x)$
To clarify with an example, if $X$ has a normal distribution and $f(x)=e^x$ then $Y=f(x)$ is a log-normal distribution. The median of $Y$ is $e^\mu$ which is a nice result. However, the standard deviation of $Y$ is $\sqrt{(\exp(\sigma ^{2})-1)\exp(2\mu +\sigma ^{2})}$ which is not a nice result. Thus, the standard deviation doesn't have this property I'm looking for.
Am I barking up the wrong tree? A strictly increasing transform is clearly relevant for measuring medians but is it stupid of me to think that this type of transform would be relevant for spread? Perhaps the second derivative of the transform is more relevant for the measure of spread.
 A: You are looking for a functional $s$ that maps a random variable $X$ into a number $s(X)$ such as 
$$s(f(X))=f(s(X))\quad(*)$$
for all increasing $f$.
The fact is : any such functional $s$ has the form $s(X)=F_X^{-1}(\alpha)$ where $F_X$ the cumulative distribution of $X$ and for some fixed $\alpha$ . Your recognize the $\alpha$ quantile. Median is the special case for $\alpha=0.5$.
Proof :
$F_X(X)$ has a uniform distribution over $[0;1]$. This is because $P(F_X(X)<a)=P(X<F_X^{-1}(a))=F_X(F_X^{-1}(a))=a$. Note that this fact is useful : any variable can be transformed into a variable that has uniform distribution this way.
Now consider a functional $s$ having the property $(*)$ :
$s(f(X))=f(s(X))$ for any increasing function $f$. Call $\alpha$ the value of $s(U)$ for a variable $U$ having a uniform distribution over $[0;1]$. Then for any variable $X$ :  $s(F_X(X))=s(U)=\alpha$. Using the property $(*)$ for $F_X$ that is increasing: $F_X(s(X))=\alpha$ that is $s(X)=F_X^{-1}(\alpha)$. 
End of proof.
A quantile cannot be seen as a scale operator. Instead, the difference between two quantiles can be seen as a scale operator (like IQR). 
An interesting question is : what property instead of $(*)$ a difference between two quantiles could satisfy ?
