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How many isomorphism classes of irreducible representations of $\mathbb{Z}/2\mathbb{Z}$ are there?

What I know:
There are only two irreducible representations of $\mathbb{Z}/2\mathbb{Z}$, namely $r$ which sends both elements to $1$ and $r'$ which sends $0$ to $-1$.
Further an isomorphism between these two representations would have to be an isomorphism $A:\mathbb{C}\rightarrow\mathbb{C}$ s.t. $$A(r(g)x)=r'(g)A(x).$$ Since $\mathbb{Z}\backslash2\mathbb{Z}$ has but two elements, I will write out this condition explicitly:
$A(x)=A(r(1)x)=r'(1)A(x)=A(x)$
$A(x)=A(r(0)x)=r'(0)A(x)=-A(x)$.

Does this mean that there is no such isomorphism, i.e. $r$ and $r'$ are not isomorphic? And does that mean the number of isomorphism classes is zero?

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    You are correct that those representations are not isomorphic (assuming you mean to send $1$ to $-1$, rather than sending $0$ there). The final conclusion is not at all what this means. The number of isomorphism classes means the number of distinct (i.e. distinct up to isomorphism) such representations, not the number of isomorphisms of any sort. – 2017-02-13
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    Ohh, and it should be $\mathbb{Z}/2\mathbb{Z}$, rather than with a $\backslash$. – 2017-02-13
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    @TobiasKildetoft Does that mean then that the number of isomorphism classes is $2$? – 2017-02-13
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    Yes, precisely. – 2017-02-13

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