Every fully invariant subgroup is characteristic, but some characteristic subgroups need not be fully invariant. For example, the center of a group is characteristic but not always fully invariant. We pursue a single example.

Recall the dihedral group of order $2n$ , denoted $D_{2n}$ , can be considered as the symmetries of a regular $n$ -gon. If we consider a regular hexagon, so $n=6$ , and label the vertices counterclockwise from 1 to 6 we can then encode each symmetry as a permutation on 6 points. So a rotation by $\pi/3$ can be encoded as the permutation $\rho = (123456)$ and the reflection fixing the axis through the vertices 1 and 4 can be encoded as $\phi = (26)(35)$ . Indeed these two permutations generate a permutation group isomorphic to $D_{12}$ .

The center of a dihedral group of order $2n$ is trivial if $n$ is odd, and of order 2 if $n>2$ is even (if $n=2$ it is the entire group $D_4\cong \mathbb{Z}_2\oplus\mathbb{Z}_2$ , see the remark below). Specifically, if $\rho$ is a rotation of order $n$ , and $n=2m$ , then $\langle \rho^m\rangle$ is the center of $D_{2n}$ . (Note this is the only rotation or order 2, and in particular it is always a rotation by $\pi$ .) So when $n=6$ , the center is $\langle (14)(25)(36)\rangle$ .

Now fix $n=6$ and note the following assignment of generators determines an endomorphism $f:D_{12}\rightarrow D_{12}$ : $$ (123456)\mapsto (26)(35),\quad (26)(35)\mapsto (14)(25)(36) $$ Note that image $K:=\langle (26)(35),(14)(25)(36)\rangle\cong \mathbb{Z}_2\oplus \mathbb{Z}_2$ , as $(14)(25)(36)$ is central in $D_{12}$ and the generators of $K$ are distinct elements of order $2$ . [This can be proved with the relations of the dihedral group.]

Now the center is mapped via $f$ to the subgroup $\langle (26)(35)\rangle$ so $f(Z(D_{12}))$ is not contained in $Z(D_{12})$ proving $Z(D_{12})$ is not fully-invariant.

Of course the example applies without serious modification to the dihedral groups on $2m$ -gons, where $m>1$ is odd. Here a generally offending endomorphism may be described with a composition of maps (the first leaves the center invariant, the second swaps the basis of the image of the first thus moving the image of the center): $$ \rho\mapsto \rho^m\mapsto \phi,\qquad \phi\mapsto \phi\mapsto \rho^m $$ As $m$ is odd and the center, $\langle \rho^m\rangle$ , has order 2, it follows $\langle\rho^m\rangle$ maps to $\langle \rho^m\rangle$ under the first map, and then can be interchanged with a reflection to violate the condition of full invariance. If $m$ is even then the center lies in the kernel of the first map so no such trick can be played.