Let $V$ be a symmetric non-degenerate random variable. Now define \begin{align} f_m(x)= E\big[|V+x|^m\big]-|x|^m, \end{align}
Is the following claim true:
$f_m(x)$ is constant if and only if $m=2$.
Clear for $m=2$ we have that $f_2(x)=E[V^2]$ and is constant for all $x$.
I am interested in the other direction. To show that the function is non-constant we can show that the derivative of $f_m(x)$ is non-zero. The derivative is given by
\begin{align}
f^{\prime}(x)= m \left( E\big[|V+x|^{m-1} {\rm sign}(V+x)\big] -\text{sign}(x) |x|^{m-1} \right), x\neq 0.
\end{align}
However, I am not sure how to show that $f^{\prime}(x)$ has values such that $f^{\prime}(x)\neq 0$.