The following should be known, but I could not find an example.


Let $\kappa$ be an uncountable cardinal. Find a model $M$ of size $\kappa$ which has $\ge\kappa$ many automorphisms, but for some $m\in M$, $(M,m)$ has only the trivial automorphism (I call this element rigid on the title. Not sure if there is a standard name for it).



A model with $\kappa$ many automorphism and a rigid element.
Let $(F,+,\cdot)$ be a rigid field, and put $M=(F,+,R)$, where $R(x,y,u,v)\iff xy=uv$. Then $(M,1)$ is interdefinable with $(F,+,\cdot)$ and therefore rigid, but $M$ has $|F|$ automorphisms $x\mapsto ax$ for each $a\in F^\times$.
There exist rigid fields of any infinite cardinality, see e.g. <https://mathoverflow.net/a/61082/12705>.


Generalizing this and Eric Wofsey’s answer, if $(A,\cdot,R\_1,R\_2,\dots,F\_1,F\_2,\dots)$ is any rigid structure such that $\cdot$ is a group operation, we can take $M=(A,\cdot',R'\_1,\dots,F'\_1,\dots)$, where
\begin{gather}
\cdot'(x,y,z)=xz^{-1}y,\\
R'\_i(x\_1,\dots,x\_n,z)\iff R\_i(x\_1z^{-1},\dots,x\_nz^{-1}),\\
F'\_i(x\_1,\dots,x\_n,z)=F\_i(x\_1z^{-1},\dots,x\_nz^{-1})z.
\end{gather}
Then $M$ has automorphisms $x\mapsto xa$ for each $a\in A$, but $(M,1)$ is rigid. Elementary constructions of rigid structures (e.g., graphs) of arbitrary cardinality are easy to find, and expanding the structure with a $\cdot$ if necessary preserves rigidity.


Since all answers given so far are instances of the construction above, let me also mention an example that works in a different way. Let $F$ be a rigid field, and let $M$ be the polynomial ring $F[x]$. Since $F^\times=M^\times$ is definable in $M$, every automorphism of $M$ fixes $F$, and is uniquely determined by the image of $x$. Thus, $M$ has automorphisms $f(x)\mapsto f(ax+b)$ for $a\in F^\times$, $b\in F$, whereas $(M,x)$ is rigid.


