Prove there's only one nonabelian semidirect product $(C_2 \times C_2) \ltimes C_3$ up to isomorphism

Question

I am trying to prove that there is only one nonabelian semidirect product $(C_2 \times C_2) \ltimes C_3$ up to isomorphism. By the definition of semidirect product, I am trying to find a group homomorphism $\phi : C_2 \times C_2 \rightarrow \text{Aut}(C_3)$. Initially, I wanted to prove this via cases. Note that $\text{Aut}(C_3) = \{\text{id}, (1 \ 2)\}$, where $(1 \ 2)$ is the permutation that switches $1$ and $2$ and id is the identity mapping. Since the trivial case produces an abelian group, the problem reduces to showing the following identifications of $\phi$ all yield isomorphic groups $(C_2 \times C_2) \ltimes C_3$ (as each produces a nonabelian semidirect product): (1) $\phi(0,1) = (1 \ 2)$, $\phi(1,0) = \text{id}$, and thus $\phi(1,1) = (1 \ 2)$; (2) $\phi(0,1) = \text{id}$, $\phi(1,0) = (1 \ 2)$, and thus $\phi(1,1) = (1 \ 2)$; and (3) $\phi(0,1) = (1 \ 2)$, $\phi(1,0) = (1 \ 2)$, and thus $\phi(1,1) = \text{id}$. However, this has become too tedious for my liking, and I'm becoming muddled in finding isomorphisms between each case. Is there a cleverer way of showing this?

Answer 1

You could argue that all those morphisms $C_2\times C_2\to \operatorname{Aut}(C_3)$ are conjugated, in the sense that for any two of them (say $f$ and $g$), there is an automorphism $\phi$ of $C_2\times C_2$ such that $f=g\circ \phi$.

This is easy to see since all those morphisms are basically the non-zero linear maps $\mathbb{F}_2^2\to \mathbb{F}_2$, so this is just basic linear algebra.

I think you can then easily establish that if those morphisms are conjugated, they induce isomorphic semi-direct products (this is true in general, for any groups).