Def1. Say $X$ is finite, if there isn't a $Y\subsetneq X$ such that $|X|=|Y|$.
Def2. Say $X$ is finite, if for every $f:X\to X$, $f$ is an injection iff $f$ is a surjection.
It can be easily verified that they are both equivalent to the usual definition, that is, $X$ is finite iff $|X|=n$ for some $n\in\mathbb N$, using axiom of choice. Also, it is known that Def1 is strictly weaker than the usual one in $ZF$. (There is an 'infinite' $X$ which satisfies Def1)
My question is, is Def1=Def2 in $ZF$, and is Def2=the usual one in $ZF$?