On automorphisms group $C_{2}\times D_{8}$ 
Let $D_{8}$ be group dihedral of order 8 and $C_{2}$  be cyclic group of order $2$.
  Then determine the number all automorphisms of $C_{2}\times D_{8}$. Can you determine automorphisms group of $C_{2}\times D_{8}$?

In general case we know that $\lvert\operatorname{Aut}(H\times K)\rvert\geqslant\lvert \operatorname{Aut}(H)\rvert\times\lvert \operatorname{Aut}(K)\rvert$ for finite groups $H$ and $K$. Also we know that if $(\lvert H\rvert, \lvert K\rvert)=1$, then $\lvert\operatorname{Aut}(H\times K)\rvert=\lvert \operatorname{Aut}(H)\rvert\times\lvert \operatorname{Aut}(K)\rvert$. If $(\lvert H\rvert, \lvert K\rvert)\neq 1$, then can we say that $\lvert\operatorname{Aut}(H\times K)\rvert>\lvert \operatorname{Aut}(H)\rvert\times\lvert \operatorname{Aut}(K)\rvert$?
Thank you
 A: This is not a complete answer.
Now $C_2 = \langle \alpha \rangle$ and $D_8 = \langle x , y\rangle$ where $\alpha$ and $x$ have order $2$ and $y$ has order $4$. Let $G = C_2 \times D_8$. For clarity, I'll denote $(\alpha, 1)$ by $\alpha$, $(1,x)$ by $x$ and $(1,y)$ by $y$ here. Because $G$ is generated by $\alpha, x$ and $y$, any automorphism $\phi$ of $G$ is completely determined by $\phi(\alpha)$, $\phi(x)$ and $\phi(y)$. 
It is not difficult to show that $y^2$ is fixed by every automorphism of $G$. Therefore since $\alpha$ is central of order $2$, there are $2$ possible choices for $\phi(\alpha)$. The element $x$ is noncentral of order $2$, so there are $8$ possible choices for $\phi(x)$. The element $y$ has order $4$ so there are $4$ possible choices for $\phi(y)$.
Thus the group $G$ has at most $2 \times 8 \times 4 = 64$ automorphisms. According to GAP, there are exactly $64$ automorphisms, so it turns out these are all the automorphisms of $G$. I haven't figured out any nice way to show that each of the possible choices for $\phi(\alpha), \phi(x)$ and $\phi(y)$ determine an automorphism. 
