Yes, there are such spaces. One example is referenced by Ramiro de la Vega in his answer to this related question.

As Joel points out in the comments, the property you describe, namely "$X$ has an anti-symmetric mappability relation", is strictly weaker than the property "$X$ is strongly rigid", where the latter is defined to mean that the only continuous functions from $X$ to itself are the identity map and the constant functions.

The paper mentioned in Ramiro's answer (available here) gives a hereditarily indecomposable continuum with this property. The construction is due to Howard Cook, and the fact that it has the desired property is expressed in Theorem 11 of his paper. (The theorem states that *the identity is the only mapping of $M_2$ onto a non-degenerate subcontinuum of $M_2$* (where $M_2$ is a continuum constructed earlier). Here *mapping* means continuous function and *non-degenerate subcontinuum* means a closed connected set having more than one point. Since the continuous image of $M_2$ is always closed and connected, this theorem implies immediately that the only non-identity continuous functions from $M_2$ to itself are the constant maps.)

mappable?--how about "$x$ pays attention to $y$" or "$x$ appreciates $y$" or .... $\endgroup$ – Włodzimierz Holsztyński Feb 2 '17 at 8:14