Semidirect product of $Z_2$ with itself

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In a previous question, I was told that $$Z_2 \rtimes_\varphi Z_2= G_2$$ $G_2$ $$\begin{array}{c|c|c|c|c} \cdot & e & a & b& c\\\hline e & e & a & b & c \\\hline a &a & e & c& b\\\hline b & b & c & a& e \\\hline c & c & b & e&a \end{array}$$ By the Wikipedia page I know that $\varphi: Z_2 \to \text{Aut}(Z_2)$
I’ve reasoned that Aut$(Z_2)$ is the trivial group.
Then, there is only one option for $\varphi$ which creates $G_1$ $$\begin{array}{c|c|c|c|c} \cdot & e & a & b& c\\\hline e & e & a & b & c \\\hline a &a & e & c& b\\\hline b & b & c & e & a \\\hline c & c & b & a & e \end{array}$$ Where am I wrong?