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I interpret your statement as being concerned with conjugation in $\mathrm{SL}_2(\mathbb Z)$. In that case I think that the arguments given only give that the groups are cyclic of order $1$, $2$, $3$, $4$ or $6$ not that they are unique up to conjugacy. For this latter fact one need only consider order $1$, $3$ or $4$ as the others are obtained by multplying a generator by $-E$. The case of order $1$ is trivial and that of order $3$ or $4$ gives a module of rank $1$ over the ring of $3$'rd and $4$'th roots of unity and then the statement is equivalent to these rings having class number $1$.
Addendum: I was a little it sketchy as Victor pointed out. In the present case all the relevant representations of $\mathbb Z[\mathbb Z/p]$factor through $\mathbb Z[\zeta_p]$ which is seen by looking at the characteristic polynomial and hence one doesn't have to look at the more general representations. In higher ranks I certainly agree with Victor about the need for the full ring. As for $\mathrm{GL}_2(\mathbb Z)$ versus $\mathrm{SL}_2(\mathbb Z)$-conjugacy I think that is taken care of by the fact that complex conjugation acts trivially on the class groups (which are trivial) and hence there is an automorphism of determinant $-1$ of the modules in question.
I interpret your statement as being concerned with conjugation in $\mathrm{SL}_2(\mathbb Z)$. In that case I think that the arguments given only give that the groups are cyclic of order $1$, $2$, $3$, $4$ or $6$ not that they are unique up to conjugacy. For this latter fact one need only consider order $1$, $3$ or $4$ as the others are obtained by multplying a generator by $-E$. The case of order $1$ is trivial and that of order $3$ or $4$ gives a module of rank $1$ over the ring of $3$'rd and $4$'th roots of unity and then the statement is equivalent to these rings having class number $1$.