I read that there are only two semidirect products $(\Bbb Z_4 \times\Bbb Z_2) \rtimes\Bbb Z_2$ whose presentations are given by
\begin{align}
(\Bbb Z_4 \times\Bbb Z_2) \rtimes\Bbb Z_2 : \langle a, b, c ~|~ a^4 = b^2 = c^2 = e, c^{-1} a c = a^3 b, c^{-1} b c = b \rangle, \\
(\Bbb Z_4 \times\Bbb Z_2) \rtimes\Bbb Z_2 : \langle a, b, c ~|~ a^4 = b^2 = c^2 = e, c^{-1} a c = a, c^{-1} b c = a^2 b \rangle.
\end{align}
However, I suspect that there are more than two possibilities to construct that.
Since the action of semidirect product is an automorphism, we can say the image of a generator under the automorphism is again a generator.
In this case, $\Bbb Z_4 \times\Bbb Z_2$ is rank two and I found that there are at least four pairs of generators as $(ab, a^3), (ab, a), (a^3 b, a), (a^3 b, a^3)$. Note that this $(*, \star)$ does not mean an element of $\Bbb Z_4 \times\Bbb Z_2$ (just a pair of generators).
Let me consider the image of $(ab, a^3)$. If we denote the automorphism as $\phi$ of $(\Bbb Z_4 \times\Bbb Z_2) \rtimes_\phi \Bbb Z_2$, I think there are eight possibilities:
\begin{align}
(\phi(ab), \phi(a^3)) = (ab, a^3), (a^3, ab), (a^3b, a), (a, a^3b), (ab, a), (a, ab), (a^3b, a^3), (a^3, a^3b).
\end{align}
Then $\phi$ maps $a, b$ to
\begin{align}
(\phi(a), \phi(b)) = (a, b), (a^3b, b), (a^3, b), (ab, b), (a^3b, a^2b), (a, a^2 b), (ab, a^2 b),
\end{align}
respectively.
Although changing the image to its inverse (e.g. $(\phi(a), \phi(b)) = (a, b) \rightarrow (a^3, b)$) gives the same group structure, there still remain four possibilities of the mapping.
Hence I think we can construct semidirect products
\begin{align}
(\Bbb Z_4 \times\Bbb Z_2) \rtimes\Bbb Z_2 : \langle a, b, c ~|~ a^4 = b^2 = c^2 = e, c^{-1} a c = a, c^{-1} b c = b \rangle, \\
(\Bbb Z_4 \times\Bbb Z_2) \rtimes\Bbb Z_2 : \langle a, b, c ~|~ a^4 = b^2 = c^2 = e, c^{-1} a c = a^3 b, c^{-1} b c = a^2 b \rangle,
\end{align}
in addition to the above two semidirect products.
Is this discussion incorrect?
I also read elements of order 4 in $(\Bbb Z_4 \times\Bbb Z_2)$ are only $a$ and $a^3 b$, but I think $ab$ and $a^3$ are also of order 4. If my understanding here is incorrect, I guess this leads the above two extra possibilities.
Best Answer
(A) In $\Bbb Z_4\times\Bbb Z_2$, there are "exactly two copies of $\Bbb Z_4$", namely $\langle a\rangle$ and $\langle ab\rangle$. So $\Bbb Z_2=\langle c\rangle$ should switch them, or normalize both.
(B) If it is switching, then, the first presentations seems to be the only possible group in semidirect product.
(C) Suppose it is normalizing both copies of $\Bbb Z_4$.
(C1) If it trivially normalizes both copies of $\Bbb Z_4$, then $G$ will be abelian.
(C2) If it acts non-trivially on only one copy of $\Bbb Z_4$, then you can check that it is second group in your list.
(C3) If it acts non-trivially on both copies of $\Bbb Z_4$, ($c^{-1}ac=a^{-1}$ and $c^{-1}bc=a^2b$), then I think, you should get a group different from these two.