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 $(0,0)$.