Automorphism inducing identity on closed points

Question

Let $k$ be an algebraically closed field. Let $X$ be a $k$-scheme that admits a $k$-immersion into projective space. Is it true that any $k$-automorphism of $X$ that induces identity on closed points is necessarily the identity?

Answer 1

If $X$ is a reduced scheme locally of finite type over an algebraically closed field $k$, a $k$-morphism $f:X\rightarrow X$ that is identity on closed points must be the identity. In particular, you don't need $f$ to be a $k$-automorphism.

The equalizer of $f$ and $\mathrm{Id}$ is a locally closed subscheme of $X$. Assumptions imply that the equalizer must contain every closed point (since $k$ is algebraically closed, every closed point comes from a unique $k$-morphism $\mathrm{Spec}\,k\rightarrow X$; and since $f$ is a $k$-morphism, the induced map on residue fields must be an isomorphism).

The complement of the equalizer of $f$ and $\mathrm{Id}$ is a constructible set containing no closed points; but in a non-empty scheme locally of finite type over an algebraically closed field closed points form a dense set, so every closed subset of $X$ has a dense set of closed points; thus every non-empty constructible set contains a closed point. Therefore, the equalizer is a closed subscheme of $X$ whose underlying space is $|X|$. Since $X$ is reduced, the equalizer is $X$ so $f=\mathrm{Id}$.

Answer 2

The simplest counterexample is given by $$ X = Spec(k[t]/t^2) $$ with the morphism given by $t \mapsto \lambda t$ (where $\lambda \in k^\times$).