There is no such functor. Recall that a split epimorphism is a morphism $f : x \to y$ with a section (right inverse) $g : y \to x$, satisfying $fg = \text{id}_y$. Split epimorphisms, as their name suggests, are epimorphisms, and moreover they are absolute epimorpisms in that they are preserved by any functor whatsoever.

In $\text{Grp}$ every split epimorphism arises as a projection $N \rtimes G \to G$ where $N \rtimes G$ is a semidirect product, so any functor from groups to groups must preserve these. In particular if $\text{Hol}(-)$ were such a functor it would follow that $\text{Hol}(N \rtimes G)$ admits a split epimorphism to $\text{Hol}(G)$.

No such split epimorphism exists in general. Explicitly, take $N = C_2, G = C_2^3$ with the trivial action. Then

$$\text{Hol}(N \rtimes G) \cong C_2^4 \rtimes GL_4(\mathbb{F}_2)$$

while

$$\text{Hol}(G) \cong C_2^3 \rtimes GL_3(\mathbb{F}_2).$$

$GL_n(\mathbb{F}_2) \cong SL_n(\mathbb{F}_2) \cong PSL_n(\mathbb{F}_2)$ is simple for $n \ge 3$, so these two groups each have a unique nonabelian simple group in their Jordan-Holder decompositions, namely $GL_4(\mathbb{F}_2)$ and $GL_3(\mathbb{F}_2)$ respectively. Any non-solvable quotient of $C_2^4 \rtimes GL_4(\mathbb{F}_2)$ must also contain $GL_4(\mathbb{F}_2)$ in its Jordan-Holder decomposition, and since $GL_3(\mathbb{F}_2)$ and $GL_4(\mathbb{F}_2)$ have different orders they are non-isomorphic, so $\text{Hol}(G)$ is not a quotient of $\text{Hol}(N \rtimes G)$.

A similar but simpler argument shows that there is no functor sending a group $G$ to its center $Z(G)$, since for example $\text{sgn} : S_3 \to C_2$ is a split epimorphism but there is no epimorphism $Z(S_3) \to Z(C_2)$ since the former is trivial and the latter is not. We can also show that there is no functor sending a group $G$ to its automorphism group $\text{Aut}(G)$ using the same counterexample $N = C_2, G = C_2^3$ as above, although $N = C_2, G = C_2^2$ also works and requires a slightly different argument. I used a similar but more complicated argument on math.SE recently to show that there is no functor sending a finite-dimensional vector space $V$ to $GL(V)$.

There is no such functor. Recall that a split epimorphism is a morphism $f : x \to y$ with a section (right inverse) $g : y \to x$, satisfying $fg = \text{id}_y$. Split epimorphisms, as their name suggests, are epimorphisms, and moreover they are absolute epimorpisms in that they are preserved by any functor whatsoever.

In $\text{Grp}$ every split epimorphism arises as a projection $N \rtimes G \to G$ where $N \rtimes G$ is a semidirect product, so any functor from groups to groups must preserve these. In particular if $\text{Hol}(-)$ were such a functor it would follow that $\text{Hol}(N \rtimes G)$ admits a split epimorphism to $\text{Hol}(G)$.

No such split epimorphism exists in general. Explicitly, take $N = C_2, G = C_2^3$ with the trivial action. Then

$$\text{Hol}(N \rtimes G) \cong C_2^4 \rtimes GL_4(\mathbb{F}_2)$$

while

$$\text{Hol}(G) \cong C_2^3 \rtimes GL_3(\mathbb{F}_2).$$

$GL_n(\mathbb{F}_2) \cong SL_n(\mathbb{F}_2) \cong PSL_n(\mathbb{F}_2)$ is simple for $n \ge 3$, so these two groups each have a unique nonabelian simple group in their Jordan-Holder decompositions, namely $GL_4(\mathbb{F}_2)$ and $GL_3(\mathbb{F}_2)$ respectively. Any non-solvable quotient of $C_2^4 \rtimes GL_4(\mathbb{F}_2)$ must also contain $GL_4(\mathbb{F}_2)$ in its Jordan-Holder decomposition, and since $GL_3(\mathbb{F}_2)$ and $GL_4(\mathbb{F}_2)$ have different orders they are non-isomorphic, so $\text{Hol}(G)$ is not a quotient (or even a subquotient) of $\text{Hol}(N \rtimes G)$.

A similar but simpler argument shows that there is no functor sending a group $G$ to its center $Z(G)$, since for example $\text{sgn} : S_3 \to C_2$ is a split epimorphism but there is no epimorphism $Z(S_3) \to Z(C_2)$ since the former is trivial and the latter is not. We can also show that there is no functor sending a group $G$ to its automorphism group $\text{Aut}(G)$ using the same counterexample $N = C_2, G = C_2^3$ as above, although $N = C_2, G = C_2^2$ also works and requires a slightly different argument. I used a similar but more complicated argument on math.SE recently to show that there is no functor sending a finite-dimensional vector space $V$ to $GL(V)$.

There is no such functor. Recall that a split epimorphism is a morphism $f : x \to y$ with a section (right inverse) $g : y \to x$, satisfying $fg = \text{id}_y$. Split epimorphisms, as their name suggests, are epimorphisms, and moreover they are absolute epimorpisms in that they are preserved by any functor whatsoever.

In $\text{Grp}$ every split epimorphism arises as a projection $N \rtimes G \to G$ where $N \rtimes G$ is a semidirect product, so any functor from groups to groups must preserve these. In particular if $\text{Hol}(-)$ were such a functor it would follow that $\text{Hol}(N \rtimes G)$ admits a split epimorphism to $\text{Hol}(G)$.

No such split epimorphism exists in general. Explicitly, take $N = C_2, G = C_2^3$ with the trivial action. Then

$$\text{Hol}(N \rtimes G) \cong C_2^4 \rtimes GL_4(\mathbb{F}_2)$$

while

$$\text{Hol}(G) \cong C_2^3 \rtimes GL_3(\mathbb{F}_2).$$

$GL_n(\mathbb{F}_2) \cong SL_n(\mathbb{F}_2) \cong PSL_n(\mathbb{F}_2)$ is simple for $n \ge 3$, so these two groups each have a unique nonabelian simple group in their Jordan-Holder decompositions, namely $GL_4(\mathbb{F}_2)$ and $GL_3(\mathbb{F}_2)$ respectively. Any non-solvable quotient of $C_2^4 \rtimes GL_4(\mathbb{F}_2)$ must also contain $GL_4(\mathbb{F}_2)$ in its Jordan-Holder decomposition, and since $GL_3(\mathbb{F}_2)$ and $GL_4(\mathbb{F}_2)$ have different orders they are non-isomorphic, so $\text{Hol}(G)$ is not a quotient of $\text{Hol}(N \rtimes G)$.

A similar but simpler argument shows that there is no functor sending a group $G$ to its center $Z(G)$, since for example $\text{sgn} : S_3 \to C_2$ is a split epimorphism but there is no epimorphism $Z(S_3) \to Z(C_2)$ since the former is trivial and the latter is not. We can also show that there is no functor sending a group $G$ to its automorphism group $\text{Aut}(G)$ using the same counterexample $N = C_2, G = C_2^3$ as above. I used a similar but more complicated argument on math.SE recently to show that there is no functor sending a finite-dimensional vector space $V$ to $GL(V)$.

There is no such functor. Recall that a split epimorphism is a morphism $f : x \to y$ with a section (right inverse) $g : y \to x$, satisfying $fg = \text{id}_y$. Split epimorphisms, as their name suggests, are epimorphisms, and moreover they are absolute epimorpisms in that they are preserved by any functor whatsoever.

In $\text{Grp}$ every split epimorphism arises as a projection $N \rtimes G \to G$ where $N \rtimes G$ is a semidirect product, so any functor from groups to groups must preserve these. In particular if $\text{Hol}(-)$ were such a functor it would follow that $\text{Hol}(N \rtimes G)$ admits a split epimorphism to $\text{Hol}(G)$.

No such split epimorphism exists in general. Explicitly, take $N = C_2, G = C_2^3$ with the trivial action. Then

$$\text{Hol}(N \rtimes G) \cong C_2^4 \rtimes GL_4(\mathbb{F}_2)$$

while

$$\text{Hol}(G) \cong C_2^3 \rtimes GL_3(\mathbb{F}_2).$$

$GL_n(\mathbb{F}_2) \cong SL_n(\mathbb{F}_2) \cong PSL_n(\mathbb{F}_2)$ is simple for $n \ge 3$, so these two groups each have a unique nonabelian simple group in their Jordan-Holder decompositions, namely $GL_4(\mathbb{F}_2)$ and $GL_3(\mathbb{F}_2)$ respectively. Any non-solvable quotient of $C_2^4 \rtimes GL_4(\mathbb{F}_2)$ must also contain $GL_4(\mathbb{F}_2)$ in its Jordan-Holder decomposition, and since $GL_3(\mathbb{F}_2)$ and $GL_4(\mathbb{F}_2)$ have different orders they are non-isomorphic, so $\text{Hol}(G)$ is not a quotient of $\text{Hol}(N \rtimes G)$.

A similar but simpler argument shows that there is no functor sending a group $G$ to its center $Z(G)$, since for example $\text{sgn} : S_3 \to C_2$ is a split epimorphism but there is no epimorphism $Z(S_3) \to Z(C_2)$ since the former is trivial and the latter is not. We can also show that there is no functor sending a group $G$ to its automorphism group $\text{Aut}(G)$ using the same counterexample $N = C_2, G = C_2^3$ as above, although $N = C_2, G = C_2^2$ also works and requires a slightly different argument. I used a similar but more complicated argument on math.SE recently to show that there is no functor sending a finite-dimensional vector space $V$ to $GL(V)$.