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.