Automorphism of a reduced scheme determined by its restriction to a dense open?

Question

Let $X$ be a reduced scheme, and let $U\subset X$ be a dense open, and let $Aut_U(X)$ be the group of automorphisms of $X$ leaving $U$ invariant (ie, which restricts to an action on $U$).

Is the natural map $Aut_U(X)\rightarrow Aut(U)$ injective?

Answer 1

No. For instance, let $X$ be $\mathbb{A}^1$ with a doubled origin and $U$ be the complement of the two origins. Then $X$ has an automorphism which is the identity on $U$ but swaps the two origins.

On the other hand, if $X$ is separated, then it is true. More generally, if $X$ is reduced and $Y$ is separated, then two morphisms $f,g:X\to Y$ which agree on a dense open subscheme $U$ of $X$ must be equal. To prove this, consider the map $F=(f,g):X\to Y\times Y$. Since $Y$ is separated, the diagonal $\Delta:X\to Y\times Y$ is a closed immersion. The pullback of $\Delta$ with respect to $F$ then is a closed immersion $Z\to X$, whose image contains $U$ since $f$ and $g$ agree on $U$. Since $U$ is dense, this means $Z\to X$ is surjective, and hence an isomorphism since $X$ is reduced. But by definition of $Z$, $f$ and $g$ agree on $Z$, so $f=g$.