Square map is endomorphism not implies abelian for loop

Statement
It is possible to have a loop $$(L,*)$$ in which the square map $$x \mapsto x^2$$ (where $$x^2$$ is the product of $$x$$ with itself) is an endomorphism but where $$L$$ is not a commutative loop, i.e., there exist $$x,y \in L$$ such that $$x * y = y * x$$.

Opposite facts

 * Square map is endomorphism iff abelian (for group)
 * Square map is endomorphism iff abelian for diassociative loop

Similar facts

 * Square map is endomorphism not implies abelian for monoid

Proof
We can construct an example non-abelian loop $$L$$ that has order five and where every element squares to the identity. Clearly, here, the square map is an endomorphism. The loop has the following multiplication table: