Give $\mathbb{R}/\mathbb{Z}\simeq S^{1}$ a the standard cyclic ordering. In other words, we put a ternary relation $C$ on $\mathbb{R}/\mathbb{Z}\simeq S^{1}$ where $x,y,z\in C$ if and only if $x,y,z$ are distinct and oriented counterclockwise (for example $(.1,.2,.3)\in C$ but $(.3,.2,.1)\not\in C$ ). Now consider the structure, $(\mathbb{R}/\mathbb{Z},C,m)$ where $m$ is the operation defined by $m(x,y,z)=(y-x)+(z-x)+x$. Consider the mapping $\phi_{a}$ where $\phi_{a}(x)=x+a$. Then $\phi_{a}$ is an automorphism of the structure $(\mathbb{R}/\mathbb{Z},C,m)$.

On the other hand, the structure $(\mathbb{R}/\mathbb{Z},C,m,0)$ has no non-trivial automorphisms: the operation $+$ is definable in the structure $(\mathbb{R}/\mathbb{Z},C,m,0)$. Furthermore, each rational number in $\mathbb{R}/\mathbb{Z}$ is easily seen to be definable in $(S^{1},C,m,0)$. From the cyclic ordering, one can therefore define each interval $[r,s]$ with rational endpoints $r,s$ in terms of a first order formula. Therefore each element in $\mathbb{R}/\mathbb{Z}$ is definable in terms of countably many first order formulas in the structure $(\mathbb{R}/\mathbb{Z},C,m,0)$. Therefore the structure $(\mathbb{R}/\mathbb{Z},C,m,0)$ has no non-trivial automorphisms. This example also works for cardinalities below the continuum simply by taking subgroups of $\mathbb{R}/\mathbb{Z}$ instead.

I claim that one can modify a counterexample with $\kappa$ non-logical symbols to obtain a counterexample with only countably many non-logical symbols. It is easy to construct structures with many relation symbols or function symbols and with many automorphisms, but where if we fix an element, then the new structure would be rigid (as was noted by Eric Wofsey). For example, if $F$ is a field and $f_{a}:F\rightarrow F$ is the mapping defined by $f_{a}(x)=ax$, then $(F,(f_{a})_{a\in A})$

Suppose that $(X,(R_{\alpha,n})_{\alpha<\kappa,n\in\mathbb{N}})$ is a structure where each $R_{\alpha,n}$ is an $n$-ary relation. Suppose now that $0\in X$ and $(X,0,(R_{\alpha,n})_{\alpha<\kappa,n\in\mathbb{N}})$ has trivial automorphism group. Then for each $n$, let $R_{n}$ be the $n$-ary relation on $X\times\kappa$ where $((x_{1},\alpha_{1}),...,(x_{n},\alpha_{n}))\in R_{n}$ if and only if $\alpha_{1}=...=\alpha_{n}$ and $(x_{1},...,x_{n})\in R_{\alpha_{1},n}$. Let $\leq$ be the partial ordering on $X\times\kappa$ where $(x,\alpha)\leq(y,\beta)$ if and only if $\alpha\leq\beta$.

Suppose $f:X\rightarrow X$ is an automorphism of $(X,(R_{\alpha,n})_{\alpha,n})$. Then define $\phi:X\times\kappa\rightarrow X\times\kappa$ by letting $\phi(x,\alpha)=(f(x),\alpha).$ Then $\phi$ is an automorphism of $(X\times\kappa,(R_{n})_{n\in\mathbb{N}},\leq)$.

Now suppose that $\phi:X\times\kappa\rightarrow X\times\kappa$ is an automorphism of $(X\times\kappa,(R_{n})_{n\in\mathbb{N}},\leq)$. Then it is easy to see that if $\phi(x,\alpha)=(y,\beta)$, then $\alpha=\beta$. Therefore for each $\alpha<\kappa$, there is some $f_{\alpha}:X\rightarrow X$ such that $\phi(x,\alpha)=(f_{\alpha}(x),\alpha)$. However, if $\alpha<\beta$, then $(x,\alpha)\leq(x,\beta)$, so $(f_{\alpha}(x),\alpha)=\phi(x,\alpha)\leq\phi(x,\beta)=(f_{\beta}(x),\beta)$. Therefore, we have $f_{\alpha}(x)=f_{\beta}(x)$. We conclude that there is some $f:X\rightarrow X$ with $\phi(x,\alpha)=(f(x),\alpha)$. Furthermore, the automorphism $f$ fixes $0$ if and only if the automorphism $\phi$ fixes $(0,0)$.

$(x_{1},...,x_{n})\in R_{\alpha,n}$ if and only if $((x_{1},\alpha),...,(x_{n},\alpha))\in R_{\alpha}$ if and only if $((f(x_{1}),\alpha),...,(f(x_{n}),\alpha))=(\phi(x_{1},\alpha),...,\phi(x_{n},\alpha))\in R_{\alpha}$ if and only if $(f(x_{1}),...,f(x_{n}))\in R_{\alpha}$. Therefore, the function $f$ is an automorphism of $(X,(R_{\alpha,n})_{\alpha,n})$. We therefore have the automorphism group of $(X,(R_{\alpha,n})_{\alpha,n})$ be in a one-to-one correspondence with the automorphism group of $(X\times\kappa,(R_{n})_{n\in\mathbb{N}},\leq)$. Furthermore, since only one automorphism of $X$ fixes $0$, only the trivial automorphism of $(X\times\kappa,(R_{n})_{n\in\mathbb{N}},\leq)$ fixes $(0,0)$.