Let \(\varphi\) be Euler's totient function.
For \(n \geq 1\), a cyclic group \(C_n\) has exactly \(\varphi(n)\) generators, and for primes \(p \neq q\) and \(k, \ell \geq 1\), \[ \begin{aligned} \varphi(p) &= p-1, \\ \varphi(p^k) &= p^{k-1}(p - 1), \\ \varphi(p^k q^\ell) &= \varphi(p^k)\,\varphi(q^\ell). \end{aligned} \]
\(\Aut(C_n) \cong \unitsof{(\ZZ/n\ZZ)}\), a group of order \(\varphi(n)\). It is cyclic if and only if \(n = 1, 2, 4, p^k\), or \(2p^k\) with \(p\) an odd prime and \(k \geq 1\).
For \(p\) an odd prime and \(k \geq 1\), \(\Aut(C_{p^k}) \cong C_{\varphi(p^k)}\); in particular \(\Aut(C_p) \cong C_{p-1}\).
\(\Aut(C_2) = 1\), \(\Aut(C_4) \cong C_2\), and for \(k \geq 3\), \(\Aut(C_{2^k}) \cong C_2 \times C_{2^{k-2}}\).
If \(G\) and \(H\) are finite groups of coprime orders, then \(\Aut(G \times H) \cong \Aut(G) \times \Aut(H)\). Consequently, if \(n = \prod_{k=1}^\ell p_k^{n_k}\) with distinct primes \(p_k\), \(p_1 = 2\), and \(n_1 \geq 3\), then \(C_n \cong \prod_{k=1}^{\ell} C_{p_k^{n_k}}\) and \[ \begin{aligned} \Aut(C_n) &\cong \prod_{k=1}^\ell \Aut\qty{C_{p_k^{n_k}}} \\ &\cong \prod_{k=1}^\ell \unitsof{(\ZZ/p_k^{n_k}\ZZ)} \\ &\cong \qty{C_2 \times C_{2^{n_1-2}} } \times \prod_{k=2}^\ell C_{m_k}, \qquad m_k \coloneqq \varphi(p_k^{n_k}) = p_k^{n_k-1}(p_k-1). \end{aligned} \]
For \(m \geq 2\) and \(n \geq 1\), \(\Aut(C_m^n) \cong \GL_n(\ZZ/m\ZZ)\). For \(p\) prime, \(\Aut(C_p^n) \cong \GL_n(\FF_p)\), and \[ \abs{\GL_n(\FF_p)} = \prod_{k=0}^{n-1}(p^n-p^k) = (p^n-1)(p^n-p)(p^n-p^2)\cdots(p^n-p^{n-1}). \]
For \(m, n \geq 1\), \(\abs{\Hom(C_n, C_m)} = \gcd(n, m)\).
Let \(N, H\) be groups, \(\psi\colon H \to \Aut(N)\) a homomorphism, \(\sigma \in \Aut(H)\), and \(\tau \in \Aut(N)\), and let \(c_\tau\colon \Aut(N) \to \Aut(N)\) be conjugation \(\alpha \mapsto \tau\alpha\inverseof{\tau}\). Then \(N \semidirect_\psi H \cong N \semidirect_{c_\tau \circ \psi \circ \sigma} H\).
For every group \(G\), \(\Inn(G) \cong G/Z(G)\).
\(\Aut(C_8) \cong \unitsof{(\ZZ/8\ZZ)} = \theset{1, 3, 5, 7}\) has every nonidentity element of order \(2\), so \(\Aut(C_8) \cong C_2 \times C_2 \not\cong C_4\).
By (h), for \(N = C_p^n\) and \(H = C_k = \gens{h}\), a homomorphism \(\psi\colon C_k \to \GL_n(\FF_p)\) is determined by the matrix \(A = \psi(h)\), and replacing \(A\) by a similar matrix \(PA\inverseof{P}\) gives an isomorphic semidirect product. So it suffices to take \(A\) in a canonical form, such as the rational canonical form.