Compute $|\operatorname{Aut} (\mathbb{(Z}/{1155\mathbb Z)}^\times)|$.
I feel puzzled about this problem. I found that $$\mathbb{(Z}/{1155\mathbb Z})^\times \cong \mathbb{(Z}/3\mathbb Z)^\times \times \mathbb{(Z}/5\mathbb Z)^\times \times \mathbb{(Z}/7\mathbb Z)^\times \times \mathbb{(Z}/{11\mathbb Z)}^\times. $$ I get the order of itself $\left|\mathbb{(Z}/{1155\mathbb Z)}^\times\right|$, but I don't know how to find its automorphism.
Hints:
We have $(\mathbb{Z}/p\mathbb{Z})^*\cong \mathbb{Z}/(p-1)\mathbb{Z}$, see here.
We know Aut $\mathbb Z/n\mathbb{Z}$, see here, and this duplicate.
You have obtained already, that $p=3,5,7,11$.
Jyrki's objection for the invariant factors, i.e., to write the direct product of these four cyclic groups in a canonical form.
Let us first rewrite the abelian group $G=\Bbb{Z}_{1155}^*$ in terms of its invariant factors. You already deduced that $$ \Bbb{Z}_{1155}^*\simeq \Bbb{Z}_3^*\times\Bbb{Z}_5^*\times\Bbb{Z}_7^*\times\Bbb{Z}_{11}^*. $$ We know that for a prime $p$, $\Bbb{Z}_p^*$ is cyclic of order $p-1$. Therefore $$ G\simeq C_2\times C_4\times C_6\times C_{10}. $$ Recall the rule that if $\gcd(m,n)=1$, then $C_m\times C_n\simeq C_{mn}$. Here $C_6\simeq C_2\times C_3$ and $C_{10}\simeq C_2\times C_5$. Regrouping gives thus $$ \begin{aligned} G&\simeq C_2\times C_4\times(C_2\times C_3)\times(C_2\times C_5)\\ &\simeq (C_4\times C_3\times C_5)\times C_2\times C_2\times C_2\\ &\simeq C_{60}\times C_2\times C_2\times C_2. \end{aligned} $$ IMHO it is easier to count the number of automorphisms of $G$ from this. Let $c_1,c_2,c_3,c_4$ be generators for the four factors, so $c_1$ is of order $60$ and $c_2,c_3,c_4$ are all of order two. We get all the elements of $G$ uniquely in the form $\prod_{i=1}^4 c_i^{a_i}$ with $0\le a_1<60$, $a_2,a_3,a_4\in\{0,1\}$.
Let $\sigma$ be an automorphism of $G$. Then obviously $\sigma(c_1)$ is of order $60$. Therefore we must have $\sigma(c_1)=c_1^{a_1}c_2^{a_2}c_3^{a_3}c_4^{a_4}$ with $\gcd(a_1,60)=1$ but no constraints on the other exponents. There are $\phi(60)=2\cdot(3-1)\cdot(5-1)=16$ choices for $a_1$, and two choices for the other exponents - a total of $16\cdot2^3=128$ choices for $\sigma(c_1)$ (we shall shortly see that all these choices lead to automorphisms of $\sigma$). Let us record the fact that because $a_1$ is necessarily odd we have $\phi(c_1^{30})=c_1^{30}$. Furthermore, this is the only element of order two in $\langle \sigma(c_1)\rangle$ (a cyclic group of an even order has a unique element of order two).
The elements of order two in $G$ are exactly the fifteen non-trivial elements of the elementary 2-abelian subgroup $H$ generated by $\{c_1^{30},c_2,c_3,c_4\}$. Of these $c_1^{30}=\sigma(c_1^{30})$ has already been used. This means that fourteen possible choices remain for $\sigma(c_2)$.
From this point the calculation proceeds very similarly to the calculation of the number of invertible matrices modulo two. Namely, $\sigma(c_3)$ can be any element of $G$ that is of order two, but is not in the subgroup generated by $\sigma(c_1^{30})$ and $\sigma(c_2)$ (otherwise $\sigma$ won't be injective). The latter condition rules out $4$ of the $16$ elements of $H$, and, given $\sigma(c_1)$ and $\sigma(c_2)$, we have $12$ choices for $\sigma(c_3)$.
Repeating the above calculation/argument we see that we have $8$ choices for $\sigma(c_4)$, namely the elements of $H\setminus\langle \sigma(c_1^{30})=c_1^{30},\sigma(c_2),\sigma(c_3)\rangle$. We easily see that all these combinations of choices lead to an injective homomorphism from $G$ to itself, hence to an automorphism. So, finally:
The number of automorphisms of the group $\Bbb{Z}_{1155}^*$ is $$|Aut(G)|=128\cdot14\cdot12\cdot8=2^{13}\cdot3\cdot7.$$
I won't give you the structure of $Aut(G)$ (I don't know how to do that concisely). I do want to point out that $Aut(G)$ obviously has a subgroup isomorphic to the simple group $GL_3(\Bbb{F}_2)$ of order $168$. In particular, $Aut(G)$ is not a solvable group.
