Showing an automorphism I am trying to show that $Aut(Z_4+Z_2)$ is the dihedral group $D_4$ or the Quaternions $Q$. The Quaternions have 1 element of order 2, while $D_4$ has 5 elements of order 2. So if I show two automorphisms, then its $D_4$. So far I have defined one automorphism as $\phi_1: (1,0)\rightarrow (3,0)$ and $(0,1)\rightarrow (0,1).$ This works because $\phi_1 \circ \phi_1=e.$ I need another one, but I cant seem to find it. I have tried sending $(1,0)$ to its other generators, $(1,1),(3,1), and (1,0)$, but the only one that works is $(1,0)$, but I know that that one is too simple as in obvious. Is it really the only one that works?
$$\phi_2(1,0):= (1,0)\;,\;\;\phi_2(0,1):=(3,1)$$

$$\phi_2^k(1,0)=(1,0)\;\forall\,k\in\Bbb N\;,\;\;$$

$$\begin{align}&amp;\phi_2^2(0,1)=\phi_2(3,1)=\phi_2\left(3(1,0)+(0,1)\right)=(3,0)+(3,1)=(2,1)\\
&amp;\phi_2^3(0,1)=\phi_2(2,1)=\phi_2\left(2(1,0)+(0,1)\right)=(2,0)+(3,1)=(1,1)\\
&amp;\phi_2^4(0,1)=\phi_2(1,1)=\phi_2\left((1,0)+(0,1)\right)=(1,0)+(3,1)=(0,1)\end{align}$$

and thus ord$(\phi_2)=4\;$

Finally

$$\begin{align}&amp;(\phi_1\phi_2\phi_1)(1,0)=\phi_1\phi_2(3,0)=\phi_1(3,0)=\phi_1\left(3(1,0)\right)=(1,0)=\phi_2^{-1}(1,0)\\
&amp;(\phi_1\phi_2\phi_1)(0,1)=\phi_1\phi_2(0,1)=\phi_1(3,1)=\phi_1\left(3(1,0)+(0,1)\right)=(1,1)=\phi_2^{-1}(0,1)\end{align}$$

So that $\;\phi_1\phi_2\phi_1=\phi_2^{-1}\;$ .Does this ring a bell?
$$\phi_2(1,0):= (1,0)\;,\;\;\phi_2(0,1):=(3,1)$$

$$\phi_2^k(1,0)=(1,0)\;\forall\,k\in\Bbb N\;,\;\;$$

$$\begin{align}&amp;\phi_2^2(0,1)=\phi_2(3,1)=\phi_2\left(3(1,0)+(0,1)\right)=(3,0)+(3,1)=(2,1)\\
&amp;\phi_2^3(0,1)=\phi_2(2,1)=\phi_2\left(2(1,0)+(0,1)\right)=(2,0)+(3,1)=(1,1)\\
&amp;\phi_2^4(0,1)=\phi_2(1,1)=\phi_2\left((1,0)+(0,1)\right)=(1,0)+(3,1)=(0,1)\end{align}$$

and thus ord$(\phi_2)=4\;$

Finally

$$\begin{align}&amp;(\phi_1\phi_2\phi_1)(1,0)=\phi_1\phi_2(3,0)=\phi_1(3,0)=\phi_1\left(3(1,0)\right)=(1,0)=\phi_2^{-1}(1,0)\\
&amp;(\phi_1\phi_2\phi_1)(0,1)=\phi_1\phi_2(0,1)=\phi_1(3,1)=\phi_1\left(3(1,0)+(0,1)\right)=(1,1)=\phi_2^{-1}(0,1)\end{align}$$

So that $\;\phi_1\phi_2\phi_1=\phi_2^{-1}\;$ .Does this ring a bell?