I assume that the r.v. $X$ integrable (hence also the copy $Y$).

The only case of equality is when the distribution of $X$ is carried by at most two points.

Indeed, by independence of $X$ and $Y$, $$E\big[|X-Y|\big|X\big] = f(X) \text{ where } f(x) := E[|x-Y|] \ge |E[x-Y]| = |x|.$$ Hence $$f(X) = E\big[|X-Y|\big|X\big] \ge |X| \text{ a.s.},$$ Looking at the expectations, one deduces that
$$E[|X-Y|] = |X| \iff f(X) \ge |X| \text{ a.s.}.$$ $$E[|X-Y|] = |X| \iff \text{ for $P_X$ a.e. } x, \quad f(x) \ge |x|.$$ $$E[|X-Y|] = |X| \iff \text{ for $P_X$ a.e. } x, E[|x-Y|] = |E[x-Y]|.$$ $$E[|X-Y|] = |X| \iff \big(\text{ for $P_X$ a.e. } x, Y-x \ge 0 \text{ a.s. or } Y-x \le 0 \text{ a.s.}\big).$$ $$E[|X-Y|] = |X| \iff \big( \text{ for $P_X$ a.e. } x, \mathrm{ess}\inf Y \ge x \text{ or } \mathrm{ess}\sup Y \le x \big).$$ 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).

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).