I am trying to prove that ${\rm Aut}(\Bbb Z_{70})$ is abelian so I am thinking of proving that it is cyclic since cyclic groups are abelian.
We know that ${\rm Aut}(\Bbb Z_{70})$ is isomorphic to $U({70})$ but also that $U(m)$ is cyclic only if $m=2,4,p,2p^r$, where $p$ is prime. $70$ cannot be written in this form so it is not cyclic right?
I'm kind of stuck here, any ideas?
Another thought of mine was to show that the order of ${\rm Aut}(\Bbb Z_{70})$ is a prime number which means that the group is cyclic but i dont know how to calculate the order of ${\rm Aut}(\Bbb Z_{70}).$
The automorphism group turns out to be $\Bbb Z_4\times \Bbb Z_6$, which is not cyclic, so your approach will not work, but it is clearly abelian. Calculating this group is, I think, most easily done with the Chinese remainder theorem, which tells you that $\Bbb Z_{70}\cong \Bbb Z_2\times\Bbb Z_5\times \Bbb Z_7$ (either as a group with only addition, or as a ring with addition and multiplication, or as a monoid with only multiplication).
Using that an element of a product ring or a product monoid is invertible iff each component is invertible, this approach ultimately gives you that the automorphism group is isomorphic to $U_2\times U_5\times U_7$, which is isomorphic to $\Bbb Z_4\times \Bbb Z_6$.
But you don't need to know exactly what the automorphism group is, how large it is, or what order its elements have. You say that know that $\operatorname{Aut}(\Bbb Z_{70})$ is isomorphic to $U_{70}$, the multiplicative group consisting of the invertible elements of $\Bbb Z_{70}$. The operation of said group is the multiplication of $\Bbb Z_{70}$, restricted to the relevant subset. Presumably you know that this multiplication is commutative on the whole of $\Bbb Z_{70}$. This almost immediately implies it is commutative on a subset.