[proofplan] We prove the conjugacy by the standard shadowing construction. For $g$ sufficiently $C^1$-close to $f$, every $f$-orbit is a small pseudo-orbit for $g$, and every $g$-orbit is a small pseudo-orbit for $f$. The shadowing lemma assigns unique nearby true orbits, while expansivity turns uniqueness of shadowing into the conjugacy relation and into inverse maps. The final homeomorphism is close to the identity because the shadowing scale is chosen arbitrarily small. [/proofplan]

[step:Choose compatible expansivity and shadowing constants] Fix a metric $d: M \times M \to [0,\infty)$ inducing the topology of $M$. Since $f$ is Anosov, it is expansive: there exists a constant $\rho_f > 0$ such that, whenever $x,y \in M$ satisfy \begin{align*} d(f^n(x), f^n(y)) \leq \rho_f \end{align*} for every $n \in \mathbb{Z}$, then $x = y$; this is the Expansivity Theorem for Anosov diffeomorphisms (citing a result not yet in the wiki: Expansivity of Anosov diffeomorphisms). By $C^1$-openness of the Anosov condition and uniform hyperbolic estimates near a fixed Anosov diffeomorphism, shrink to a neighbourhood $\mathcal U_1$ of $f$ and choose $\rho_*>0$ such that every $g\in\mathcal U_1$ is Anosov and expansive with expansivity scale $\rho_*$. We use the following uniform shadowing consequence on this neighbourhood: for every shadowing scale $\alpha>0$ there is a tolerance $\delta_*(\alpha)>0$ such that every two-sided $\delta_*(\alpha)$-pseudo-orbit for any $g\in\mathcal U_1$ is $\alpha$-shadowed by a true $g$-orbit. Choose a number $\eta > 0$ such that \begin{align*} 0 < 4\eta < \min\{\rho_f,\rho_*\}. \end{align*} By the Shadowing Lemma for Anosov diffeomorphisms (citing a result not yet in the wiki: Shadowing Lemma for Anosov diffeomorphisms), there exists $\delta_f > 0$ such that every two-sided $\delta_f$-pseudo-orbit for $f$ is $\eta$-shadowed by a true $f$-orbit. That is, if $(x_n)_{n \in \mathbb{Z}}$ is a sequence in $M$ satisfying \begin{align*} d(f(x_n), x_{n+1}) < \delta_f \end{align*} for every $n \in \mathbb{Z}$, then there exists $y \in M$ such that \begin{align*} d(f^n(y), x_n) < \eta \end{align*} for every $n \in \mathbb{Z}$. Because $4\eta < \rho_f$, such a shadowing point is unique. Indeed, if $y,z \in M$ both $\eta$-shadow the same pseudo-orbit $(x_n)_{n \in \mathbb{Z}}$, then for every $n \in \mathbb{Z}$ the triangle inequality gives \begin{align*} d(f^n(y), f^n(z)) \leq d(f^n(y), x_n) + d(x_n, f^n(z)) < 2\eta < \rho_f. \end{align*} Expansivity of $f$ implies $y=z$. Set $\delta_*:=\delta_*(\eta)$ for the uniform shadowing tolerance supplied above. [/step]

[step:Choose the perturbation neighbourhood so true orbits become pseudo-orbits]Since $M$ is compact and the $C^1$ topology is stronger than the $C^0$ topology, there exists a $C^1$ neighbourhood $\mathcal{U}_0$ of $f$ in $\operatorname{Diff}^1(M)$ such that every $g \in \mathcal{U}_0$ satisfies \begin{align*} \sup_{x \in M} d(g(x), f(x)) < \min\{\delta_f,\delta_*\}. \end{align*} Replace $\mathcal U_0$ by $\mathcal U_0\cap\mathcal U_1$ and fix $g \in \mathcal{U}_0$. For each $x \in M$, define the two-sided $g$-orbit sequence \begin{align*} \gamma_x: \mathbb{Z} \to M, \qquad n \mapsto g^n(x). \end{align*} Then $\gamma_x$ is a $\delta_f$-pseudo-orbit for $f$, since for every $n \in \mathbb{Z}$, \begin{align*} d(f(\gamma_x(n)), \gamma_x(n+1)) = d(f(g^n(x)), g(g^n(x))) < \delta_f. \end{align*} By the previous step, there is a unique point $H(x) \in M$ satisfying \begin{align*} d(f^n(H(x)), g^n(x)) < \eta \end{align*} for every $n \in \mathbb{Z}$. This defines a map \begin{align*} H: M \to M, \qquad x \mapsto H(x). \end{align*}[/step]

[guided]The goal is to assign to each $g$-orbit a genuine $f$-orbit that follows it for all positive and negative time. The compactness of $M$ lets us choose the $C^1$ neighbourhood $\mathcal{U}_0$ so small that $g$ is uniformly close to $f$ as a map: \begin{align*} \sup_{x \in M} d(g(x), f(x)) < \min\{\delta_f,\delta_*\}. \end{align*} This is the only point in this step where the topology on $\operatorname{Diff}^1(M)$ is used: $C^1$-closeness implies $C^0$-closeness. Now fix $x \in M$ and define the orbit sequence \begin{align*} \gamma_x: \mathbb{Z} \to M \end{align*} \begin{align*} n \mapsto g^n(x). \end{align*} We check that this sequence is a $\delta_f$-pseudo-orbit for $f$. For each $n \in \mathbb{Z}$, the next point in the sequence is $\gamma_x(n+1)=g(g^n(x))$, while applying $f$ to the current point gives $f(\gamma_x(n))=f(g^n(x))$. Since the uniform distance between $f$ and $g$ is less than $\delta_f$, we get \begin{align*} d(f(\gamma_x(n)), \gamma_x(n+1)) = d(f(g^n(x)), g(g^n(x))) < \delta_f. \end{align*} Thus the Shadowing Lemma for Anosov diffeomorphisms applies to $\gamma_x$. It gives a point $H(x) \in M$ whose $f$-orbit $\eta$-shadows the $g$-orbit of $x$: \begin{align*} d(f^n(H(x)), g^n(x)) < \eta \end{align*} for every $n \in \mathbb{Z}$. The uniqueness proved in the previous step makes $H(x)$ well-defined, so we obtain a map \begin{align*} H: M \to M \end{align*} \begin{align*} x \mapsto H(x). \end{align*}[/guided]

[step:Use uniqueness of shadowing to obtain the semiconjugacy] We prove \begin{align*} f \circ H = H \circ g. \end{align*} Fix $x \in M$. The point $f(H(x))$ $\eta$-shadows the $g$-orbit of $g(x)$, because for every $n \in \mathbb{Z}$, \begin{align*} d(f^n(f(H(x))), g^n(g(x))) = d(f^{n+1}(H(x)), g^{n+1}(x)) < \eta. \end{align*} By definition, $H(g(x))$ is the unique point whose $f$-orbit $\eta$-shadows the $g$-orbit of $g(x)$. Hence \begin{align*} f(H(x)) = H(g(x)). \end{align*} Since $x \in M$ was arbitrary, \begin{align*} f \circ H = H \circ g. \end{align*} [/step]

[step:Prove continuity of the shadowing map] We prove that $H: M \to M$ is continuous. Let $(x_k)_{k=1}^{\infty}$ be a sequence in $M$ converging to $x \in M$. Since $M$ is compact, every subsequence of $(H(x_k))_{k=1}^{\infty}$ has a further subsequence converging to some point $y \in M$. It is enough to prove $y=H(x)$ for every such subsequential limit. Fix a subsequence, still denoted $(x_k)_{k=1}^{\infty}$, such that $H(x_k) \to y$. For each fixed $n \in \mathbb{Z}$, continuity of the diffeomorphisms $f^n: M \to M$ and $g^n: M \to M$ gives \begin{align*} f^n(H(x_k)) \to f^n(y) \end{align*} and \begin{align*} g^n(x_k) \to g^n(x). \end{align*} Since \begin{align*} d(f^n(H(x_k)), g^n(x_k)) < \eta \end{align*} for every $k$ and every $n \in \mathbb{Z}$, passing to the limit gives \begin{align*} d(f^n(y), g^n(x)) \leq \eta \end{align*} for every $n \in \mathbb{Z}$. The defining estimate for $H(x)$ gives \begin{align*} d(f^n(H(x)), g^n(x)) < \eta \end{align*} for every $n \in \mathbb{Z}$. Hence \begin{align*} d(f^n(y),f^n(H(x)))<2\eta<\rho_f \end{align*} for every $n\in\mathbb Z$. Expansivity of $f$ gives $y=H(x)$. Therefore every subsequential limit of $H(x_k)$ equals $H(x)$, and compactness implies $H(x_k) \to H(x)$. Hence $H$ is continuous. [/step]

[step:Construct the inverse shadowing map] Because $g\in\mathcal U_0\subset\mathcal U_1$, the uniform constants $\rho_*$ and $\delta_*$ chosen at the beginning apply to $g$. Since \begin{align*} \sup_{x \in M} d(g(x), f(x)) < \delta_*, \end{align*} every $f$-orbit is a $\delta_*$-pseudo-orbit for $g$. Applying the uniform shadowing theorem for $g$ gives a continuous map \begin{align*} K: M \to M \end{align*} such that \begin{align*} g \circ K = K \circ f \end{align*} and \begin{align*} d(g^n(K(y)), f^n(y)) < \eta \end{align*} for every $y \in M$ and every $n \in \mathbb{Z}$. We show that $K$ and $H$ are inverse maps. Fix $x \in M$. Using the relation $g \circ K = K \circ f$, for every $n \in \mathbb{Z}$ we have \begin{align*} g^n(K(H(x))) = K(f^n(H(x))). \end{align*} More directly, the defining estimates give, for every $n\in\mathbb Z$, \begin{align*} d(g^n(K(H(x))),f^n(H(x)))<\eta \end{align*} and \begin{align*} d(f^n(H(x)),g^n(x))<\eta. \end{align*} Therefore \begin{align*} d(g^n(K(H(x))),g^n(x))<2\eta<\rho_*. \end{align*} Expansivity of $g$ with scale $\rho_*$ gives \begin{align*} K(H(x)) = x. \end{align*} Similarly, for every $y\in M$ and every $n\in\mathbb Z$, the estimates \begin{align*} d(f^n(H(K(y))),g^n(K(y)))<\eta \end{align*} and \begin{align*} d(g^n(K(y)),f^n(y))<\eta \end{align*} give \begin{align*} d(f^n(H(K(y))),f^n(y))<2\eta<\rho_f. \end{align*} Expansivity of $f$ gives \begin{align*} H(K(y)) = y \end{align*} for every $y \in M$. Thus $H$ is a bijection with inverse $K$. Since both $H$ and $K$ are continuous, $H$ is a homeomorphism. [/step]

[step:Convert to the stated conjugacy convention and control closeness to the identity] The construction gives \begin{align*} f \circ H = H \circ g. \end{align*} Let \begin{align*} h: M \to M, \qquad y \mapsto H^{-1}(y). \end{align*} Then $h$ is a homeomorphism and, composing the identity $f \circ H = H \circ g$ on the left and right by $H^{-1}$, we obtain \begin{align*} h \circ f = g \circ h. \end{align*} Finally, the shadowing estimate at time $n=0$ gives \begin{align*} d(H(x), x) < \eta \end{align*} for every $x \in M$. Since $\eta>0$ was chosen arbitrarily subject to the expansivity and shadowing constraints, the initial $C^1$ neighbourhood of $f$ may be shrunk so that $H$, and hence its inverse $h=H^{-1}$, lies in any prescribed $C^0$ neighbourhood of $\operatorname{id}_M$. Therefore $f$ is $C^1$ structurally stable. [/step]