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 $f$.