4 Clarification

The automorphism group functor is not representable in general. Consider for instance the case of an algebraically closed field $k$ and $\mathbf A^2=\mathrm{Spec}\ k[x,y]$. Assume the automorphism group functor for this scheme is representable by an algebraic space $Aut$. For each $n$ we have an automorphism $(x,y)\mapsto(x+\sum_{1\leq i\leq n}t^iy^i,y)$ over $k[t]/(t^{n+1})$ giving a set of compatible morphisms $\mathrm{Spec}k[t]/(t^{n+1})\to Aut$. That system comes from a morphism $\mathrm{Spec}k[[t]]\to \mathrm{Spec}\ k[[t]]\to Aut$ (see Emerton's response to this question), which works also for maps into algebraic spaces at least as $k$ is algebraically closed). However, the corresponding automorphism would have to be $(x,y)\mapsto(x+\sum_{1\leq i}t^iy^i,y)$ which doesn't make sense.

3 Typo fix

The automorphism group functor is not representable in general. Consider for instance the case of an algebraically closed field $k$ and $\mathbf A^2=\mathrm{Spec}\ k[x,y]$. Assume the automorphism group functor for this scheme is representable by an algebraic space $Aut$. For each $n$ we have an automorphism $(x,y)\mapsto(x+\sum_{1\leq i\leq n}t^iy^i,y)$ over $k[t]/(t^{n+1})$ giving a set of compatible morphisms $\mathrm{Spec}k[t]/(t^{n+1})\to Aut$. That system comes from a morphism $\mathrm{Spec}k[[t]]\to Aut$ (see Emerton's response to this question). However, the corresponding automorphism would have to be $(x,y)\mapsto(x+\sum_{1\leq i}t^iy^i,y)$ which doesn't make sense.

2 added 2 characters in body

The automorphism group functor is not representable in general. Consider for instance the case of an algebraically closed field $k$ and $\mathbf A^2=\mathrm{Spec}k{x,y}$A^2=\mathrm{Spec}\ k[x,y]$. Assume the automorphism group functor for this scheme is representable by an algebraic space$Aut$. For each$n$we have an automorphism$(x,y)\mapsto(x+\sum_{1\leq i\leq n}t^iy^i,y)$over$k[t]/(t^{n+1})$giving a set of compatible morphisms$\mathrm{Spec}k[t]/(t^{n+1})\to Aut$. That system comes from a morphism$\mathrm{Spec}k[[t]]\to Aut$(see Emerton's response to this question. However, the corresponding automorphism would have to be$(x,y)\mapsto(x+\sum_{1\leq i}t^iy^i,y)\$ which doesn't make sense.

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