Every automorphism of $k((x))$ preserves $k[[x]]$. This argument is adapted from an [answer of Will Sawin][1]. Let $V$ be the set of valuations $v : k((x))^{\times} \to \mathbb{Z}$ which are $0$ on $k^{\times}$. As usual, we put $v(0) = \infty$ for any valuation $v$. I claim that $f \in k[[x]]$ if and only if $v(f) \geq 0$ for all $v \in V$.

Clearly, if $f \not\in k[[x]]$, then $v(f)<0$ for the standard valuation $v$.

In the other direction, let $v \in V$. Choose $n$ relatively prime to the characteristic of $k$. Let $f$ be of the form $1+\sum_{geq 1} a_j x^j$, then $f$ has an $n^j$-th root in $k((x))$ for all $j>0$. So $n^j | v(f)$ and we deduce that $v(f)=0$ for such an $f$. Any $g \in k[[x]]$ is the sum of such an $f$ and an element of $k$, so any such $g$ has $v(g) \geq 0$.


  [1]: https://mathoverflow.net/a/322689/297