Let $$\begin{align} \phi_x:G & \longrightarrow G\\ g & \longmapsto \phi_x(g)=x^{-1}gx \end{align}$$ be the $x$-inner automorphism.
If $G$ is a group and $H\le G$ is a characteristic subgroup then $H\unlhd G$ because $\phi_x(H)\subseteq H$ for every $x\in G$ and $h\in H\implies h=x^{-1}(xhx^{-1})x=\phi_x(xhx^{-1})\in\phi_x(H)$ because $xhx^{-1}\in\phi_{x^{-1}}(H)\subseteq H$ so we are done.
Thinking about this, does $\phi_x(H)\subseteq H$ for some particular $x\in G$ imply $\phi_x(H)=H$? The "$\phi_{x^{-1}}(H)\subseteq H$" step seems necessary so I don't think the implication is true. If the group is finite then the implication holds Are there any simple examples where it fails?
Here's a relatively simple one. Let $\Bbb{Q}^\times$ act on $\Bbb{Q}$ by multiplication, then $\Bbb{Q}\rtimes \Bbb{Q}^\times$ gives a counterexample, with the subgroup $H=\Bbb{Z}\times \{0\}$ and $x=(0,1/2)$. Then $x^{-1}Hx = 2H\subsetneq H$.
$%(q,\alpha)\cdot (r,\beta) = (q+r\alpha,\alpha\beta)$ More explicitly, we can identify the group with the subgroup of $\operatorname{GL}_2(\Bbb{Q})$ of the form $$\newcommand\bmat{\begin{pmatrix}}\newcommand\emat{\end{pmatrix}}\bmat \alpha & q\\0 & 1\emat, %\bmat \beta & r \\ 0 & 1 \emat$$ with $\alpha \in \Bbb{Q}^\times$ and $q \in \Bbb{Q}$.
Then the subgroup is $\bmat 1 & n \\ 0 & 1 \emat$, with $n\in\Bbb{Z}$, and we want to conjugate by $\bmat 1/2 & 0 \\ 0 & 1 \emat$.
Explicitly conjugating, we get $$\bmat 2 & 0 \\ 0 & 1 \emat \bmat 1 & n \\ 0 & 1 \emat \bmat 1/2 & 0 \\ 0 & 1 \emat = \bmat 1 & 2n \\ 0 & 1 \emat.$$