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).