Suppose $G$ is a group. Then, a holomorph of $G$ is defined as a outer semidirect product ${\rm Hol}(G) = G \rtimes_{\phi} {\rm Aut}(G)$. Since an outer semidirect product depends on a automorphism $\phi \in {\rm Aut}G$, I thought it cannot be determined uniquely. However, I see the Klein four-group is given by \begin{align} {\rm Hol}(C_8) = C_8 \rtimes (C_2 \times C_2) \end{align} with its presentation being \begin{align} \langle a, b, c \mid a^8 = b^2 = c^2 =e , b^{-1}ab = a^{-1}, c^{-1}a{c} = a^5\rangle. \end{align}
I don't understand well but, does this mean that we fixed $\phi$ by denoting the explicit presentation?
In this case, $\phi$ is set as a conjugacy, but still the result power of $a$ e.g. $b^{-1}ab = a^{-1}$ is dependent on choice, I think. Is my comprehension correct?
If so, it would be great if you could show an concrete example showing non-uniqueness of a holomorph.