Isomorphism types of semidirect products $\mathbb Z/n\mathbb Z\rtimes\mathbb Z/2\mathbb Z$ 
Let $n = p_1 p_2 \cdots p_k$ be the product of pairwise distinct odd primes. Let $X$ be the set of element in $\operatorname{Aut}(\mathbb Z/n\mathbb Z)$ of order $1$ or $2$. For each $\psi\in X$, the mapping
  $$\varphi_\psi : \mathbb Z/2\mathbb Z \to \operatorname{Aut}(\mathbb Z/n\mathbb Z),\quad x\mapsto \begin{cases}\operatorname{id} & \text{if }x = 0 \\ \psi & \text{if }x = 1\end{cases}$$
  is a homomorphism of groups.
  So each $\psi\in X$ defines a semidirect product
  $$G_\psi = \mathbb Z/n\mathbb Z\rtimes_{\varphi_\psi} \mathbb Z/2\mathbb Z.$$
  Are these groups $G_\psi$ pairwise non-isomorphic?

This question came to my mind when giving this answer on the classification of groups of order $30$.
We have
\begin{align*}\operatorname{Aut}(\mathbb Z/n\mathbb Z)
& \cong (\mathbb Z/n\mathbb Z)^\times \\
& \cong (\mathbb Z/p_1\mathbb Z)^\times \times \ldots \times (\mathbb Z/p_k \mathbb Z)^\times \\
& \cong \mathbb Z/(p_1 - 1)\mathbb Z\times \cdots \times \mathbb Z/(p_k - 1)\mathbb Z.\end{align*}
In the latter representation, the set $X$ is given by
$$X = \left\{(a_1,\ldots,a_k) \mathrel{}\middle|\mathrel{} a_i\in\left\{0, \frac{p_i - 1}{2}\right\}\right\}$$
So $\lvert X\rvert = 2^k$.
It is clear that $G_\psi$ is abelian (even cyclic) if and only if $\psi = \operatorname{id}$.
So for $k = 1$, the answer is yes. In the case $n = 3\cdot 5$ (which leads to the classification of the groups of order 30) the answer is yes as well, and I vaguely conjecture that the answer should be yes in general. So probably we should look at some kind of an invariant which separates the groups $G_\psi$.
 A: For any such $G$, and each $i$, the element $\frac{n}{p_i} \in \mathbb{Z}/n\mathbb{Z}$, together with the generator of $\mathbb{Z}/2\mathbb{Z}$, generates a subgroup $G_i$ of the form $\mathbb{Z}/p_i\mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}$.
I claim that $\varphi_\psi$ can be reconstructed once you know the isomorphism type of each $G_i$.
If you believe my claim, then this is all you need, because there are only two possible semidirect products $\mathbb{Z}/p_i\mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}$: an abelian one and a non-abelian one.
Here's a little push towards seeing why this claim is true: the isomorphism type of $G_i$ is determined precisely by $\varphi_\psi (1) (\frac{n}{p_i})$.
A: For each of the $2^k - 1$ divisors $d\neq 1$ of $n$, we define the group $$H_d = D_d \times \mathbb Z/(n/d)\mathbb Z.$$
The groups $H_d$ are non-abelian groups of order $2n$, and because of
$$
D_d \times \mathbb Z/(n/d)\mathbb Z \\ \cong (\mathbb Z/d\mathbb Z \rtimes \mathbb Z/2\mathbb Z) \times \mathbb Z/(n/d)\mathbb Z\\ \cong (\mathbb Z/d\mathbb Z\times \mathbb Z/(n/d)\mathbb Z) \rtimes \mathbb Z/2\mathbb Z\\ \cong \mathbb Z/n\mathbb Z \rtimes \mathbb Z/2\mathbb Z
$$
the groups are of the form considered in the question.
Since $d$ is odd, there number of elements of order $1$ or $2$ in $D_d$ is $d$.
Furthermore, since $n/d$ is odd, there are no elements of order $2$ in $Z/(n/d)\mathbb Z$.
This shows that the number of elements of order $2$ in $H_d$ is $d$. So the groups $H_d$ are pairwise non-isomorphic.
Adding the abelian group $\mathbb Z/2n\mathbb Z$, we see that there are at least $2^k$ isomorphism types of semidirect products of the form $\mathbb Z/n\mathbb Z \rtimes \mathbb Z/2\mathbb Z$.
Since there are only $2^k$ possibilities to define a homomorphism $\mathbb Z/2\mathbb Z \to \operatorname{Aut}(\mathbb Z/n\mathbb Z)$ (see the question), we get that the number of isomorphism types of semidirect products $\mathbb Z/n\mathbb Z \rtimes \mathbb Z/2\mathbb Z$ is exactly $2^k$, and those given in the question are pairwise non-isomorphic.
