# Arbitrary automorphism function on Aut(Quaternion Group)?

Let $\phi$ be an arbitrary automorphism of $Aut(Q_8)$. I need to prove that $\phi (i) \in \{i, -i, j, -j, k, -k\}$.

Can someone explain why $\phi(i)$ could not be 1 or -1, even if $\phi$ is technically arbitrary? I don't understand why just because $\phi$ is an automorphism that those two elements of the quaternion group would be excluded/why $\phi (i)$ could not equal 1 or -1.

Automorphisms (or more generally injective group homomorphisms) preserve order. $\{i, -i, j, -j, k, -k\}$ is precisely the set of elements of order $4$ in $Q_8$.
To give more details, suppose that $\phi(i)=1$, then $\phi$ is not injective, because we also have $\phi(1)=1$, so $\phi$ can't be an automorphism.
Suppose that $\phi(i)=-1$, then $\phi(-1)=\phi(i^2)=\phi(i)\phi(i)=(-1)^2=1$, so as $\phi(1)=1$, $\phi$ is not injective.