I encountered this counting question and the question goes as such:
How many automorphisms are there in $\mathbb{Z}_{105}$ of order $12$, meaning that $T(T(....(x)....))$, $n$ times yields back $x$ (that is the identity automorphism).
I began by considering the generator $1$ for $\mathbb{Z}_{105}$ and getting the following results. If $T(1) = a$, then this completely determines the automorphism, so for some $x$, we know that $T(x) = T(1x) = T(1) + T(1) + .... + T(1) = xT(1)$, meaning that $T$ can defined as: $$T(x), x \rightarrow xa \mod 105$$
Taking the automorphism of this again, then we know that $T(T(x)) = a(ax) \mod 105$, so then we require that $T$ maps $1$ to some $a$, such that: $$\begin{equation}\begin{aligned}a^nx &\equiv x \mod 105\\\implies a^n&\equiv 1 \mod 105\end{aligned}\end{equation}$$
This is of course, not easy to determine and after looking at the textbook answer, the solution is this:
We can answer this question more easily by asking how many elements of order $12$ there are in $\mathbb{Z}_2 \oplus \mathbb{Z}_{4} \oplus \mathbb{Z}_6$, and there are $16$ of them, so there are 16 automorphisms of order $12$
I have no idea how they got $\mathbb{Z}_2 \oplus \mathbb{Z}_{4} \oplus \mathbb{Z}_6$ and what the correlation of that is to the automorphisms. Perhaps $\phi(105) = 48$? Since we know that $\text{Aut}(105) \approx \text{U}(105)$? Could someone please enlighten me on this solution or another way of solving this?
As you already realised at the end, every $\phi \in \text{Aut}(105)$ is represented by multiplication with a unit mod 105. As $105 = 3 \cdot 5 \cdot7$, by the Chinese Remainder Theorem $\mathbb{Z}_{105} \cong \mathbb{Z}_3 \bigoplus \mathbb{Z}_5 \bigoplus \mathbb{Z}_7$ with units $U(3) \bigoplus U(5) \bigoplus(7) \cong \mathbb{Z}_2 \bigoplus \mathbb{Z}_4 \bigoplus \mathbb{Z}_6$