Let $X$ be an infinite compact Hausdorff space and let $\sigma\colon X\to X$ be a minimal homeomorphism thereof. Then $\sigma$ gives rise to an automorphism $\sigma'$ of $C(X)$ defined by $\sigma'(f):=f\circ\sigma^{-1}$. While reading a paper, I then come across a line that says that there is a unitary $u$ such that $ufu^{*}=\sigma'(f)$ for all $f\in C(X)$.
Is $u\in C(X)$? And how does one obtain such a unitary?
The crossed product is a construction that implements a group action on a $C^\ast$-algebra as an action by unitaries. Similar to group $C^\ast$-algebras, there are two variants, the reduced and the full (universal) crossed product.
Let $A$ be a $C^\ast$-algebra, $G$ a (discrete) group and $\alpha\colon G\longrightarrow \mathrm{Aut}(A)$ a group homomorphism. The full crossed product $A\rtimes_\alpha G$ is the universal $C^\ast$-algebra generated by $A$ and $C^\ast(G)$ subject to the relation $u_g a u_g^\ast=\alpha_g(a)$.
The reduced crossed product is constructed as follows. Assume for simplicity that $A\subset B(H)$. Define representations of $A$ and $G$ on $H\otimes\ell^2(G)$ by \begin{align*} &\pi\colon A\longrightarrow B(H\otimes \ell^2(G)),\,\pi(a)(\xi\otimes \delta_g)=\alpha_{g^{-1}}(a)\xi\otimes\delta_g\\ &\lambda\colon G\longrightarrow B(H\otimes\ell^2(G)),\,\lambda(h)(\xi\otimes\delta_g)=\xi\otimes \delta_{hg} \end{align*} The reduced crossed product $A\rtimes_{\alpha,\mathrm{r}}G$ is the $C^\ast$-algebra generated by $\pi(A)$ and $\lambda(G)$.
The elements $\pi(a)$ and $\lambda(g)$ satisfiy the relation $\lambda(g)\pi(a)\lambda(g)^\ast=\pi(\alpha_{g}(a))$. Thus there exists a (surjective) $\ast$-homomorphism $\Lambda\colon A\rtimes_\alpha G\longrightarrow A\rtimes_{\alpha,\mathrm{r}} G$ such that $\Lambda(a)=\pi(a)$ and $\Lambda(u_g)=\lambda(g)$.
In your example, $A=C(X)$, $G=\mathbb{Z}$, $\alpha(n)=(\sigma')^n$, and $C^\ast(\mathbb{Z})$ is the universal $C^\ast$-algebra generated by a unitary $u_1$. Since $\mathbb{Z}$ is amenable, the map $\Lambda$ is injective, that is, the full and reduced crossed product coincide.
What is even more, if the homeomorphism $\sigma$ is not only minimal, but also free, then the crossed product $A\rtimes G$ is simple. Thus whenever you have representation $\pi'$ of $C(X)$ on a Hilbert space $K$ and a unitary $u\in B(K)$ such that $u\pi'(f)u^\ast=\pi'(\sigma'(f))$, there exists a unique $\ast$-isomorphism from $C(X)\rtimes \mathbb{Z}$ onto the $C^\ast$-algebra generated by $\pi'(C(X))$ and $u$ that is the identity on $C(X)$ and maps $u_1$ to $u$.
All these statements should be contained in Dana Williams's book on crossed products of $C^\ast$-algebras.