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.

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.

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 $f$.