I'll just point out an answer based on somewhat different criteria than explicitly knowing the hyperbolic metric: As is well-known, every oriented, compact hyperbolic surface $C$ is canonically an algebraic curve of genus $g\ge2$.

As Ian points out, if $C$ is hyperelliptic, it always has a symmetry, the hyperelliptic involution $\iota:C\to C$, and it can be written, almost canonically, as a plane curve $$ y^2 = (x-\lambda_1)\cdots(x-\lambda_{2g+2}), $$ where the $\lambda_i$ are distinct complex numbers. Let $\Gamma\subset\mathrm{PSL}(2,\mathbb{C})$ be the (finite) group of linear fractional transformations that preserve the set $\Lambda =\{\lambda_1,\ldots,\lambda_{2g+2}\}$. Then the group of orientation preserving isometries of $C$ will be an extension of $\Gamma$ by a $\mathbb{Z}_2$ (because of the hyperelliptic involution). If one sets $\Gamma_+$ to be the (possibly slightly larger) group of $\pm$-holomorphic linear fractional transformations of $\mathbb{P}^1$ that preserves $\Lambda$, then the full group of isometries of $C$ will be a $\mathbb{Z}_2$ extension of $\Gamma_+$.

If $C$ is not hyperelliptic, then its canonical mapping $\kappa:C\to\mathbb{CP}^{g-1}$ is an embedding, and the isometries of $C$ are exactly the $\pm$-holomorphic automorphisms of $\mathbb{CP}^{g-1}$ that preserve the image $\kappa(C)$. For example, when $g=3$ and $C$ is not hyperelliptic, the image $\kappa(C)\subset\mathbb{CP}^2$ is a nonsingular plane quartic $Q(X,Y,Z)=0$, and every nonsingular plane quartic is a canonical curve of genus $3$. The orientation-preserving isometries of $C$ are exactly the elements of $\mathrm{PSL}(3,\mathbb{C})$ that preserve the quartic $Q$. Since one can write down explicit homogeneous nonsingular quartics in $3$ variables that have no nontrivial automorphisms, it follows that the generic curve of genus $3$ has no nontrivial automorphisms, and hence the corresponding hyperbolic metric has no nontrivial symmetries.

This gives one a way to write down 'explicit' examples, modulo, of course, the fact that there is no known way to write down the explicit hyperbolic metric on a nonhyperelliptic curve of genus $3$.

I'll just point out an answer based on somewhat different criteria than explicitly knowing features of the hyperbolic metric: As is well-known, every oriented, compact hyperbolic surface $C$ is canonically a Riemann surface of genus $g\ge2$.

As Ian points out, if $C$ is hyperelliptic, it always has a symmetry, the hyperelliptic involution $\iota:C\to C$, and it can be written, almost canonically, as an algebraic plane curve $$ y^2 = (x-\lambda_1)\cdots(x-\lambda_{2g+2}), $$ where the $\lambda_i$ are distinct complex numbers. Let $\Gamma\subset\mathrm{PSL}(2,\mathbb{C})$ be the (finite) group of linear fractional transformations that preserve the set $\Lambda =\{\lambda_1,\ldots,\lambda_{2g+2}\}$. Then the group of orientation preserving isometries of $C$ will be an extension of $\Gamma$ by a $\mathbb{Z}_2$ (because of the hyperelliptic involution). If lets $\Gamma_+$ be the (possibly slightly larger) group of $\pm$-holomorphic linear fractional transformations of $\mathbb{P}^1$ that preserve $\Lambda$, then the full group of isometries of $C$ will be a $\mathbb{Z}_2$ extension of $\Gamma_+$.

If $C$ is not hyperelliptic, then its canonical mapping $\kappa:C\to\mathbb{CP}^{g-1}$ is an embedding, and the isometries of $C$ are exactly the $\pm$-holomorphic automorphisms of $\mathbb{CP}^{g-1}$ that preserve the image $\kappa(C)$. For example, when $g=3$ and $C$ is not hyperelliptic, the image $\kappa(C)\subset\mathbb{CP}^2$ is a nonsingular plane quartic $Q(X,Y,Z)=0$, and every nonsingular plane quartic is a canonical curve of genus $3$. The orientation-preserving isometries of $C$ are exactly the elements of $\mathrm{PSL}(3,\mathbb{C})$ that preserve the quartic $Q$. Since one can write down explicit homogeneous nonsingular quartics in $3$ variables that have no nontrivial automorphisms, it follows that the generic curve of genus $3$ has no nontrivial automorphisms, and hence the conformal hyperbolic metric on $C$ has no nontrivial isometries.

This gives one a way to write down 'explicit' examples, modulo, of course, the fact that there is no known way explicitly to 'write down' the conformal hyperbolic metric on a nonhyperelliptic curve of genus $3$.