A subgroup $C$ of a group $G$ is said to be a characteristic subgroup of $G$ if and only if $T[C] \subset C$ for all automorphisms $T$ of $G$.
Here $T[C] \colon= \{ T(c) \colon c \in C \}$ is the image of set $C$ under the mapping $T$.
I can show that every characteristic subgroup of $G$ is a normal subgroup of $G$.
Does the converse hold too? If so, how to prove it? If not, what counter-example can be given?
On
Converse is true when $Inn(G)=Aut(G)$,
as an example it true for $S_4$ as $Inn(S_4)= Aut(S_4)\cong S_4$ and $A_4$ and $V_4$ are both normal and characteristic subgroup of $S_4$.
In general it is not true, being characteristic is stronger than being normal.
Smallest such example $G=K_4\cong Z_2 \times Z_2$, even if every subgroup is normal, it has no nontrivial characteristic subgroup.
The converse is not true. See Wikipedia for examples and a nice brief discussion.