Since the distribution of the r.v. $X$ is zero-mean, $X$ integrable (hence also the copy $Y$).
The equality is attained iff the distribution of $X$ is carried by at most two points.
Indeed, by independence of $X$ and $Y$,
$$E\big(|X-Y|\big|X\big) = f(X) \text{ where } f(x) := E|x-Y|.$$
Hence
$$f(X) =E\big(|X-Y|\big|X\big) \ge |X| \text{ a.s.}$$
Looking at the expectations, one deduces that
\begin{equation}
\begin{aligned}
E|X-Y| = E|X| &\iff f(X) = |X| \text{ a.s.} \\
&\iff \text{ for $P_X$ a.e. } x, \quad f(x) = |x| \\
&\iff \text{ for $P_X$ a.e. } x, E|x-Y| = |E(x-Y)| \\
&\iff \big(\text{ for $P_X$ a.e. } x, Y-x \ge 0 \text{ a.s. or } Y-x \le 0 \text{ a.s.}\big) \\
&\iff \big( \text{ for $P_X$ a.e. } x, \mathrm{ess}\inf Y \ge x \text{ or } \mathrm{ess}\sup Y \le x \big).
\end{aligned}
\end{equation}
Since $X$ and $Y$ have the same distribution, the last condition means that the support of $P_X$ contains at most two points (namely the essential inf and the essential sup).
