Is there a 'conjugation' on every algebraically closed field? Let $K$ be an algebraically closed field. Then the polynomial $x^2+1\in K[x]$ has two distinct roots (when $K$ doesn't have characteristic 2). Let's suggestively call them $i$ and $-i$.

Does there exist an automorphism $\phi$ of $K$ such that $\phi(i)=-i$?

This is something I've been wondering about for a while. I'm asking this question here because I, unfortunately, have very little background in field theory and have no clue on how to approach this problem. Any help is appreciated.
 A: Henning's answer is correct.  However, if we let $F$ be the smallest subfield of $K$ and $f(x) = x^2 + 1$ is irreducible in $F$, then your automorphism is legitimate.  Here's why:
Let $L \subset K$ be its splitting field. Since $L$ is a splitting field of $f$, then it is by definition a Galois extension. At this point, we can consider the automorphism group $Aut(L/F)$.  An element $\phi \in Aut(L/F)$ is a $F$-automorphism.  That is to say, it is an automorphism of $L$ (that can be extended to $K$) that fixes $F$.  
Next, it is a theorem that, since $L$ is a Galois extension $[L:F] = |Aut(L/F)|$.  It is also a theorem that the elements of $Aut(L/F)$ are uniquely determined by their actions, as permutations, on the roots of the polynomial.  
Obviously, the identity automorphism will fix the roots of $f$.  Since $f$ is irreducible in $F$, then $[L:F] > 1$, and so there must exist at least one nontrivial automorphism.  Since there are only two roots that can be permuted, then that nontrivial automorphism must swap the roots of $f$.
A: Not necessarily. For example, in the algebraic closure of $\mathbb F_5$, the two roots of $x^2+1$ are $1+1$ and $1+1+1$, and there's obviously no automorphism that interchanges them.
