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.