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The answer is actually yes, and here is an example with $n=6$ (inspired by Broughton's classification quoted by Danny and by Jim's answer).

Take the group $G = \mathbb{Z}/ 6 \mathbb{Z}$ presented as follows:$$G = \langle x, \, y \; | \; x^2=y^3=1, \, [x, y]=1 \rangle.$$ By Riemann Existence Theorem, any Galois cover of $\mathbb{P}^1:=\mathbb{P}^1(\mathbb{C})$ with group $G$ is given by a finite collection of generators $g_1, \ldots, g_k \in G$ whose product is the identity. Now choose $$g_1=x, \quad g_2=x, \quad g_3=y, \quad g_4=y^2.$$

The corresponding Galois cover $\pi \colon M \to \mathbb{P}^1$ has four branch points. The points over two of them have stabilizer of order $2$ (generated by $x$) and the points over the remaining two have stabilizer of order $3$ (generated by $y$).

By Hurwitz formula we have $$2g(M)-2 = |G| \bigg(2g(\mathbb{P}^1)-2+ 2\big(1-\frac{1}{2} \big) + 2 \big(1-\frac{1}{3} \big) \bigg) = 2,$$ that is $M$ is a compact manifold of genus $2$. The automorphism of $M$ corresponding to the generator $g:=xy \in G$ has order $6$ and is fixed point-free (whereas both $g^2$ and $g^3$ have fixed points).

The answer is actually yes, and here is an example with $n=6$ (inspired by Broughton's classification quoted by Danny and by Jim's answer).

Take the group $G = \mathbb{Z}/ 6 \mathbb{Z}$ presented as follows:$$G = \langle x, \, y \; | \; x^2=y^3=1, \, [x, y]=1 \rangle.$$ By Riemann Existence Theorem, any Galois cover of $\mathbb{P}^1:=\mathbb{P}^1(\mathbb{C})$ with group $G$ is given by a finite collection of generators $g_1, \ldots, g_k \in G$ whose product is the identity. Now choose $$g_1=x, \quad g_2=x, \quad g_3=y, \quad g_4=y^2.$$

The corresponding Galois cover $\pi \colon M \to \mathbb{P}^1$ has four branch points. The points over two of them have stabilizer of order $2$ (generated by $x$) and the points over the remaining two have stabilizer of order $3$ (generated by $y$).

By Hurwitz formula we have $$2g(M)-2 = |G| \bigg(2g(\mathbb{P}^1)-2+ 2\big(1-\frac{1}{2} \big) + 2 \big(1-\frac{1}{3} \big) \bigg) = 2,$$ that is $M$ is a compact manifold of genus $2$. The automorphism of $M$ corresponding to the generator $g:=xy \in G$ has order $6$ and is fixed point-free (whereas both $g^2=y^2$ and $g^3=x$ have fixed points).

The answer is actually yes, and here is an example with $n=6$ (inspired by Broughton's classification quoted by Danny and by Robert's answer).

Take the group $G = \mathbb{Z}/ 6 \mathbb{Z}$ presented as follows:$$G = \langle x, \, y \; | \; x^2=y^3=1, \, [x, y]=1 \rangle.$$ By Riemann Existence Theorem, any Galois cover of $\mathbb{P}^1:=\mathbb{P}^1(\mathbb{C})$ with group $G$ is given by a finite collection of generators $g_1, \ldots, g_k \in G$ whose product is the identity. Now choose $$g_1=x, \quad g_2=x, \quad g_3=y, \quad g_4=y^2.$$

The corresponding Galois cover $\pi \colon M \to \mathbb{P}^1$ has four branch points. The points over two of them have stabilizers of order $2$ (generated by $x$) and the points over the remeining two have stabilizers of order $3$ (generated by $y$).

By Hurwitz formula we have $$2g(M)-2 = |G| \bigg(2g(\mathbb{P}^1)-2+ 2\big(1-\frac{1}{2} \big) + 2 \big(1-\frac{1}{3} \big) \bigg) = 2,$$ that is $M$ is a compact manifold of genus $2$. The automorphism of $M$ corresponding to the generator $g:=xy \in G$ has order $6$ and is fixed point-free (whereas both $g^2$ and $g^3$ have fixed points).

The answer is actually yes, and here is an example with $n=6$ (inspired by Broughton's classification quoted by Danny and by Jim's answer).

Take the group $G = \mathbb{Z}/ 6 \mathbb{Z}$ presented as follows:$$G = \langle x, \, y \; | \; x^2=y^3=1, \, [x, y]=1 \rangle.$$ By Riemann Existence Theorem, any Galois cover of $\mathbb{P}^1:=\mathbb{P}^1(\mathbb{C})$ with group $G$ is given by a finite collection of generators $g_1, \ldots, g_k \in G$ whose product is the identity. Now choose $$g_1=x, \quad g_2=x, \quad g_3=y, \quad g_4=y^2.$$

The corresponding Galois cover $\pi \colon M \to \mathbb{P}^1$ has four branch points. The points over two of them have stabilizer of order $2$ (generated by $x$) and the points over the remaining two have stabilizer of order $3$ (generated by $y$).

By Hurwitz formula we have $$2g(M)-2 = |G| \bigg(2g(\mathbb{P}^1)-2+ 2\big(1-\frac{1}{2} \big) + 2 \big(1-\frac{1}{3} \big) \bigg) = 2,$$ that is $M$ is a compact manifold of genus $2$. The automorphism of $M$ corresponding to the generator $g:=xy \in G$ has order $6$ and is fixed point-free (whereas both $g^2$ and $g^3$ have fixed points).

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The answer is actually yes, and here is an example with $n=6$ (inspired by Broughton's classification quoted by Danny and by Robert's answer).

Take the group $G = \mathbb{Z}/ 6 \mathbb{Z}$ presented as follows:$$G = \langle x, \, y \; | \; x^2=y^3=1, \, [x, y]=1 \rangle.$$ By Riemann Existence Theorem, any Galois cover of $\mathbb{P}^1:=\mathbb{P}^1(\mathbb{C})$ with group $G$ is given by a finite collection of generators $g_1, \ldots, g_k \in G$ whose product is the identity. Now choose $$g_1=x, \quad g_2=x, \quad g_3=y, \quad g_4=y^2.$$

The corresponding Galois cover $\pi \colon M \to \mathbb{P}^1$ has four branch points. The points over two of them have stabilizers of order $2$ (generated by $x$) and the points over the remeining two have stabilizers of order $3$ (generated by $y$).

By Hurwitz formula we have $$2g(M)-2 = |G| \bigg(2g(\mathbb{P}^1)-2+ 2\big(1-\frac{1}{2} \big) + 2 \big(1-\frac{1}{3} \big) \bigg) = 2,$$ that is $M$ is a compact manifold of genus $2$. The automorphism of $M$ corresponding to the generator $g:=xy \in G$ has order $6$ and is fixed point-free (whereas both $g^2$ and $g^3$ have fixed points).