Groups with all normal subgroups characteristic Today in my research, I had to use fairly explicitly the rather tautological property of finite cyclic groups that every normal subgroup is characteristic, i.e. fixed by all automorphisms. This got me wondering:

do (finite) groups with the property that every normal subgroup is characteristic
have a name and/or can they be completely
classified? Generally, has this
property been investigated at all?

Apart from cyclic groups, some groups possessing the above property that immediately come to mind are simple groups, symmetric groups, semi-dihedral groups, and dihedral groups of twice odd order (however not of twice even order).
This is as far as I got on my short walk home (apart from some false claims, see comments). I suspect that this property might be well studied.
Edit:
The reference that Beren Sanders provided in his answer and the references to and from it all deal with $p$-groups. I still haven't been able to find anything about arbitrary finite groups. Some of the questions that $p$-group theorists ask are just not terribly interesting in the case of arbitrary groups. E.g. the paper that Beren Sanders mentioned proves that every finite $p$-group is contained in another finite $p$-group in which every normal subgroup is characteristic. The same statement for arbitrary finite groups is trivial: just embed your group into a symmetric group. I would still be surprised if nobody had tried to say something reasonably general about finite groups with this property.
 A: Here is a small contribution. If no two factors in a composition series for $G$ are isomorphic, then every normal subgroup of $G$ is characteristic. In fact, in groups of this type, no two distinct normal subgroups can be isomorphic. To see this, suppose $M$ and $N$ are distinct and isomorphic. Look at a composition series obtained by refining the series $1 \subseteq M \cap N < M <MN \subseteq G$. Let $S$ be a simple group occurring as a composition factor between $M \cap N$ and $M$. Then $S$ is not isomorphic to a composition factor of $M \cap N$, but it is a composition factor of $M$ and hence also of the isomorphic group $N$. Since $S$ is a composition factor of $N$ but not of $M \cap N$, it is a composition factor of $N/(N \cap M)$. Since this group is isomorphic to $MN/M$, we see that $MN/M$ has $S$ as a composition factor. This is a contradiction because then $G$ has two composition factors isomorphic to $S$: one between $M \cap N$ and $M$ and another between $M$ and $MN$.
A: 
The question in which $p$-groups all normal subgroups are characteristic is a fairly old problem in the theory of finite $p$-groups. It seems difficult to assess because of the fact that neither subgroups nor factors of a group which has none but characteristic normal subgroups need retain this property.

These are the first two sentences of the following 2009 paper published in the Israel Journal of Mathematics:
B. Wilkens. $p$-groups without noncharacteristic normal subgroups. Isr. J. Math. 172, 357–369 (2009). Zbl 1188.20014
As a non-expert I have nothing more to add.
