[proofplan] We prove both directions using empirical measures. If $\mu$ is the unique invariant probability measure and a continuous test function has non-uniform averages, then a sequence of empirical measures has a weak-$*$ cluster point whose invariance follows from the telescoping boundary term; uniqueness forces that cluster point to be $\mu$, contradicting the bad averages. Conversely, uniform convergence first implies that $\mu$ is invariant by comparing averages of $f$ and $f\circ T$, and then any invariant probability measure integrates every continuous $f$ the same way as $\mu$, so it must equal $\mu$. [/proofplan] [step:Use empirical measures to prove uniform convergence from unique ergodicity] Assume that $\mu$ is the unique $T$-invariant Borel probability measure. Let $f:X\to\mathbb{R}$ be continuous. Suppose, toward contradiction, that $A_Nf$ does not converge uniformly to the constant $\int_X f\,d\mu$. Then there exist $\varepsilon>0$, integers $N_j\to\infty$, and points $x_j\in X$ such that \begin{align*} \left| \frac{1}{N_j}\sum_{n=0}^{N_j-1}f(T^n x_j) - \int_X f\,d\mu \right| \geq \varepsilon \end{align*} for every $j\in\mathbb{N}$. For each $j$, define the empirical Borel probability measure \begin{align*} \nu_j:\mathcal{B}(X)&\to[0,1]\\ B&\mapsto \frac{1}{N_j}\sum_{n=0}^{N_j-1}\delta_{T^n x_j}(B), \end{align*} where $\delta_y$ is the Dirac probability measure at $y\in X$. Since $X$ is compact metrizable, the space $\mathcal{M}(X)$ of Borel probability measures on $X$ is weak-$*$ sequentially compact. Passing to a subsequence, still denoted $(\nu_j)$, there exists a Borel probability measure $\nu$ such that $\nu_j\to\nu$ in the weak-$*$ topology. We show $\nu$ is $T$-invariant. Let $g:X\to\mathbb{R}$ be continuous. Then \begin{align*} \int_X g\circ T\,d\nu_j-\int_X g\,d\nu_j &= \frac{1}{N_j}\sum_{n=0}^{N_j-1}g(T^{n+1}x_j) - \frac{1}{N_j}\sum_{n=0}^{N_j-1}g(T^n x_j)\\ &= \frac{g(T^{N_j}x_j)-g(x_j)}{N_j}. \end{align*} Since $g$ is continuous on compact $X$, the number $\|g\|_\infty:=\sup_{x\in X}|g(x)|$ is finite, and therefore \begin{align*} \left| \int_X g\circ T\,d\nu_j-\int_X g\,d\nu_j \right| \leq \frac{2\|g\|_\infty}{N_j}\to0. \end{align*} Taking the limit along the weak-$*$ convergent subsequence gives \begin{align*} \int_X g\circ T\,d\nu=\int_X g\,d\nu \end{align*} for every continuous $g:X\to\mathbb{R}$. Hence $T_\#\nu=\nu$. By unique ergodicity, $\nu=\mu$. But weak-$*$ convergence also gives \begin{align*} \frac{1}{N_j}\sum_{n=0}^{N_j-1}f(T^n x_j) = \int_X f\,d\nu_j \to \int_X f\,d\nu = \int_X f\,d\mu, \end{align*} contradicting the inequality with $\varepsilon$. Therefore $A_Nf\to\int_X f\,d\mu$ uniformly on $X$. [guided] Assume $\mu$ is the only invariant probability measure. If uniform convergence failed for a continuous $f$, we could find orbit segments whose empirical $f$-averages stay at least $\varepsilon$ away from $\int_X f\,d\mu$. We encode each bad orbit segment by the empirical measure \begin{align*} \nu_j=\frac{1}{N_j}\sum_{n=0}^{N_j-1}\delta_{T^n x_j}. \end{align*} Compactness of $X$ gives weak-$*$ sequential compactness of $\mathcal{M}(X)$, so after passing to a subsequence $\nu_j\to\nu$. The key point is that any weak-$*$ limit of long empirical orbit measures is invariant. For continuous $g:X\to\mathbb{R}$, \begin{align*} \int_X g\circ T\,d\nu_j-\int_X g\,d\nu_j = \frac{g(T^{N_j}x_j)-g(x_j)}{N_j}. \end{align*} The numerator is bounded by $2\|g\|_\infty$, so the difference tends to $0$. Passing to the weak-$*$ limit yields \begin{align*} \int_X g\circ T\,d\nu=\int_X g\,d\nu \end{align*} for every continuous $g$, which is the continuous-test-function characterization of $T_\#\nu=\nu$. Unique ergodicity forces $\nu=\mu$. Now test the same subsequence against the original function $f$: \begin{align*} \frac{1}{N_j}\sum_{n=0}^{N_j-1}f(T^n x_j) = \int_X f\,d\nu_j \to \int_X f\,d\nu = \int_X f\,d\mu. \end{align*} This contradicts the choice of the bad orbit segments. Hence no continuous $f$ can fail uniform convergence. [/guided] [/step] [step:Use uniform convergence to prove invariance and uniqueness] Assume condition (ii). We first prove that $\mu$ is $T$-invariant. Let $f:X\to\mathbb{R}$ be continuous. Since $f\circ T:X\to\mathbb{R}$ is continuous, condition (ii) applies to both $f$ and $f\circ T$. For every $N\geq1$ and every $x\in X$, \begin{align*} A_N(f\circ T)(x)-A_N f(x) &= \frac{1}{N}\sum_{n=0}^{N-1}f(T^{n+1}x) - \frac{1}{N}\sum_{n=0}^{N-1}f(T^n x)\\ &= \frac{f(T^N x)-f(x)}{N}. \end{align*} The right-hand side converges uniformly to $0$ because $f$ is bounded on compact $X$. Taking uniform limits in the left-hand side gives \begin{align*} \int_X f\circ T\,d\mu-\int_X f\,d\mu=0. \end{align*} Thus $T_\#\mu=\mu$. Now let $\nu$ be any $T$-invariant Borel probability measure on $X$. For every continuous $f:X\to\mathbb{R}$ and every $N\geq1$, \begin{align*} \int_X A_N f\,d\nu &= \frac{1}{N}\sum_{n=0}^{N-1}\int_X f(T^n x)\,d\nu(x)\\ &= \frac{1}{N}\sum_{n=0}^{N-1}\int_X f\,d\nu\\ &= \int_X f\,d\nu. \end{align*} Since $A_Nf$ converges uniformly to the constant $\int_X f\,d\mu$, we may pass to the limit under the integral against the probability measure $\nu$: \begin{align*} \int_X f\,d\nu = \lim_{N\to\infty}\int_X A_N f\,d\nu = \int_X f\,d\mu. \end{align*} This holds for every continuous $f:X\to\mathbb{R}$. Borel probability measures on compact metrizable spaces are determined by their integrals against continuous real-valued functions, so $\nu=\mu$. Therefore $\mu$ is the unique $T$-invariant Borel probability measure. [guided] Uniform convergence does two jobs. First, it proves that the measure appearing in the limit is actually invariant. For continuous $f$, \begin{align*} A_N(f\circ T)(x)-A_Nf(x)=\frac{f(T^Nx)-f(x)}{N}. \end{align*} The numerator is uniformly bounded, so this difference tends uniformly to $0$. But the two sequences of functions converge uniformly to the constants $\int_X f\circ T\,d\mu$ and $\int_X f\,d\mu$. Therefore these constants are equal: \begin{align*} \int_X f\circ T\,d\mu=\int_X f\,d\mu. \end{align*} This is exactly $T_\#\mu=\mu$ tested against continuous functions. Second, uniform convergence rules out every other invariant measure. Let $\nu$ be $T$-invariant. Invariance gives \begin{align*} \int_X f(T^n x)\,d\nu(x)=\int_X f\,d\nu \end{align*} for every $n\geq0$, hence \begin{align*} \int_X A_Nf\,d\nu=\int_X f\,d\nu. \end{align*} Since $A_Nf$ converges uniformly to the constant $\int_X f\,d\mu$, the integrals converge to $\int_X f\,d\mu$. Therefore $\int_X f\,d\nu=\int_X f\,d\mu$ for every continuous $f$. On a compact metrizable space, continuous functions determine Borel probability measures, so $\nu=\mu$. Thus $\mu$ is the only invariant probability measure. [/guided] [/step]