On group automorphism of subgroup a group $G$ Let $G$ be a group and $H$ be a subgroup of $G$. When is $\rm{Aut}(H)$ a subgroup of $\rm{Aut}(G)$?
The restriction of an automorphism of G to H is an automorphism of H requires that H is a characteristic  sub-groupe  in G, in this case the restriction morphism of Aut (G)   to Aut (H) has a sense and is an epimorphism. In  this cas it is hoped that Aut (G) is a direct product  of Aut (H)  ie  the exacte sequence 

I said that a sufficient condition for the $Aut(H)$
to be  injected into $Aut(G)$ is that $H$  a characteristic
subgroup in $G$ and the  exact  short sequence:  $ ker(rest)
\hookrightarrow Aut(G)\rightarrow Aut(H)$(this last morphism is
the epimorphism restriction) split. Then a partial answer that I
deduced from the last comment given to this issue as similar
question in terms of exact short sequence of groups is this: under
the condition H to be a direct factor of G and simultaneously a
 characteristic subgroup in G, then $Aut(H)$ is injected into
$Aut(G)$. This injection may in no case be canonical (as Aut (.)
Is not a functor), but the restriction is canonical and has a
sense here because H supposed characteristic.

Prove: Suppose $G=H\times K$
and $H$ characteristic in $G$, so the restriction morphism
$Aut(G)\rightarrow Aut(H)$ is an epimorphism $f\mapsto f_H $, it
can viewed in this situation as the composite morphism $p_H\circ
f\circ i_H : H\hookrightarrow H\times K \rightarrow H\times
K\times \rightarrow H $ where $i_H$ the canonical
monomorphism,$f\in Aut(H\times K)$ and  $p_H$ the canonical
projection. Since $H$ characteristic in $H\times K$ then  $K$ is
also characteristic in $H\times K$ because for any $f\in
Aut(H\times K)$ the composite morphism $p_K\circ f$ left via the
naturel projection $p_K:H\times K\rightarrow K$. and so we
obtained   the direct product $Aut(H\times K)\simeq Aut(H)\times
Aut(K)$ with the isomorphism restriction $f\mapsto (f_H,f_K)$ and
of course the already exact short sequence split.
The restriction of an automorphism of G to H is an automorphism of H requires that H is a characteristic  sub-groupe  in G, in this case the restriction morphism of Aut (G)   to Aut (H) has a sense and is an epimorphism. In  this cas it is hoped that Aut (G) is a direct product  of Aut (H)  ie  the exacte sequence 

I said that a sufficient condition for the $Aut(H)$
to be  injected into $Aut(G)$ is that $H$  a characteristic
subgroup in $G$ and the  exact  short sequence:  $ ker(rest)
\hookrightarrow Aut(G)\rightarrow Aut(H)$(this last morphism is
the epimorphism restriction) split. Then a partial answer that I
deduced from the last comment given to this issue as similar
question in terms of exact short sequence of groups is this: under
the condition H to be a direct factor of G and simultaneously a
 characteristic subgroup in G, then $Aut(H)$ is injected into
$Aut(G)$. This injection may in no case be canonical (as Aut (.)
Is not a functor), but the restriction is canonical and has a
sense here because H supposed characteristic.

Prove: Suppose $G=H\times K$
and $H$ characteristic in $G$, so the restriction morphism
$Aut(G)\rightarrow Aut(H)$ is an epimorphism $f\mapsto f_H $, it
can viewed in this situation as the composite morphism $p_H\circ
f\circ i_H : H\hookrightarrow H\times K \rightarrow H\times
K\times \rightarrow H $ where $i_H$ the canonical
monomorphism,$f\in Aut(H\times K)$ and  $p_H$ the canonical
projection. Since $H$ characteristic in $H\times K$ then  $K$ is
also characteristic in $H\times K$ because for any $f\in
Aut(H\times K)$ the composite morphism $p_K\circ f$ left via the
naturel projection $p_K:H\times K\rightarrow K$. and so we
obtained   the direct product $Aut(H\times K)\simeq Aut(H)\times
Aut(K)$ with the isomorphism restriction $f\mapsto (f_H,f_K)$ and
of course the already exact short sequence split.