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Is it consistent in $mathsfZF$ that there is a set $X$ with more than $1$ point such that every injective map $f:Xto X$ has a fixed point?

As Yair suggests, a strongly amorphous set has this property.

Recall that an amorphous set is a set which cannot be split into two infinite sets. A strongly amorphous set is a set such that in addition to being amorphous, every partition has only finitely many non-singletons. While we didn't require that it is an infinite set, this is the common assumption, so the existence of an amorphous set contradicts the axiom of choice.

One can easily show that an amorphous set cannot be linearly ordered, and that a strongly amorphous set cannot carry a group structure.

Now. If $X$ is strongly amorphous and and $fcolon Xto X$ is a permutation (which it has to be if $f$ is injective, since amorphous sets are Dedekind-finite), then the orbits of $f$ are all finite and form a partition of $X$. Therefore all but finitely many points are moved at all.

One could argue that perhaps this is too extreme of an example. Well, in Cohen's first model with a Dedekind-finite set of reals, the canonical Dedekind-finite set (that of the generic Cohen reals) is linearly ordered, but has the property that any partition into finite parts is almost entirely singletons (i.e. all but finitely many parts are singletons).

The same argument now works on that set as well. If you have a permutation, its orbits must be finite, so almost all of them are singletons.

One can also have analogues of strongly amorphous sets for larger cardinals. For example an $aleph_1$-amorphous set is a set where every subset is countable or co-countable. Being strongly $aleph_1$-amorphous means that all but countably many parts of any partition are singletons.

And it is worth noting that $sf DC_kappa$ is consistent with the existence of a $kappa^+$-amorphous set. So you can't use $sf DC$ principles to avoid this issue.