There are examples involving the $p$-adics: for instance $\mathbb{Q}_p$ itself has trivial automorphism group. Indeed as $\mathbb{Q}(i)$ embeds inside $\mathbb{Q}_p$ when $p\equiv3$ (mod 4) then $\mathbb{Q}(i)$ does embed inside a field with trivial automorphism group. Indeed this is the case for all number fields (finite extensions of $\mathbb{Q}$).

Now for an example of an algebraically closed field, $K$, an extension $L$ of $K$ and an automorphism $\tau$ of $K$ not extending to one of $L$. Let $K$ be the algebraic closure of $\mathbb{Q}$, considered as a subfield of $\mathbb{C}$ and let $\tau$ be complex conjugation. Let $L=K(x,\sqrt{x^3+ax+b})$ be the function field of an elliptic curve $E$ over $K$. Each automorphism of $L$ takes $K$ to itself. Suppose the $j$-invariant of $E$ is $i$ (considered as an element of $K$). Then any automorphism of $L$ taking $K$ to itself must fix $i$, and so cannot restrict to $\tau$ on $K$.

There are examples involving the $p$-adics: for instance $\mathbb{Q}_p$ itself has trivial automorphism group. Indeed as $\mathbb{Q}(i)$ embeds in $\mathbb{Q}_p$ when $p\equiv1$ (mod 4) then $\mathbb{Q}(i)$ does embed in a field with trivial automorphism group. Indeed this is the case for all number fields (finite extensions of $\mathbb{Q}$).

Now for an example of an algebraically closed field, $K$, an extension $L$ of $K$ and an automorphism $\tau$ of $K$ not extending to one of $L$. Let $K$ be the algebraic closure of $\mathbb{Q}$, considered as a subfield of $\mathbb{C}$ and let $\tau$ be complex conjugation. Let $L=K(x,\sqrt{x^3+ax+b})$ be the function field of an elliptic curve $E$ over $K$. Each automorphism of $L$ takes $K$ to itself. Suppose the $j$-invariant of $E$ is $i$ (considered as an element of $K$). Then any automorphism of $L$ taking $K$ to itself must fix $i$, and so cannot restrict to $\tau$ on $K$.

There are examples involving the $p$-adics: for instance $\mathbb{Q}_p$ itself has trivial automorphism group. Indeed as $\mathbb{Q}(i)$ embeds inside $\mathbb{Q}_p$ when $p\equiv3$ (mod 4) then $\mathbb{Q}(i)$ does embed inside a field with trivial automorphism group. Indeed this is the case for all number fields (finite extensions of $\mathbb{Q}$).

There are examples involving the $p$-adics: for instance $\mathbb{Q}_p$ itself has trivial automorphism group. Indeed as $\mathbb{Q}(i)$ embeds inside $\mathbb{Q}_p$ when $p\equiv3$ (mod 4) then $\mathbb{Q}(i)$ does embed inside a field with trivial automorphism group. Indeed this is the case for all number fields (finite extensions of $\mathbb{Q}$).

Now for an example of an algebraically closed field, $K$, an extension $L$ of $K$ and an automorphism $\tau$ of $K$ not extending to one of $L$. Let $K$ be the algebraic closure of $\mathbb{Q}$, considered as a subfield of $\mathbb{C}$ and let $\tau$ be complex conjugation. Let $L=K(x,\sqrt{x^3+ax+b})$ be the function field of an elliptic curve $E$ over $K$. Each automorphism of $L$ takes $K$ to itself. Suppose the $j$-invariant of $E$ is $i$ (considered as an element of $K$). Then any automorphism of $L$ taking $K$ to itself must fix $i$, and so cannot restrict to $\tau$ on $K$.