  Let $V$ be a finite-dimensional vector space over the field with $p$
elements, where $p$ is a prime number. Given arbitrary $\alpha,\beta\in
\mathrm{GL}(V)$, we consider the semidirect products $V\rtimes\langle
\alpha\rangle$ and $V\rtimes\langle \beta\rangle$, and show that if
$V\rtimes\langle \alpha\rangle$ and $V\rtimes\langle \beta\rangle$ are
isomorphic, then $\alpha$ must be similar to a power of $\beta$ that generates
the same subgroup as $\beta$; that is, if $H$ and $K$ are cyclic subgroups of
$\mathrm{GL}(V)$ such that $V\rtimes H\cong V\rtimes K$, then $H$ and $K$ must
be conjugate subgroups of $\mathrm{GL}(V)$. If we remove the cyclic condition,
there exist examples of non-isomorphic, let alone non-conjugate, subgroups $H$
and $K$ of $\mathrm{GL}(V)$ such that $V\rtimes H\cong V\rtimes K$. Even if we
require that non-cyclic subgroups $H$ and $K$ of $\mathrm{GL}(V)$ be abelian,
we may still have $V\rtimes H\cong V\rtimes K$ with $H$ and $K$ non-conjugate
in $\mathrm{GL}(V)$, but in this case, $H$ and $K$ must at least be isomorphic.
If we replace $V$ by a free module $U$ over ${\mathbf Z}/p^m{\mathbf Z}$ of
finite rank, with $m>1$, it may happen that $U\rtimes H\cong U\rtimes K$ for
non-conjugate cyclic subgroups of $\mathrm{GL}(U)$. If we completely abandon
our requirements on $V$, a sufficient criterion is given for a finite group $G$
to admit non-conjugate cyclic subgroups $H$ and $K$ of $\mathrm{Aut}(G)$ such
that $G\rtimes H\cong G\rtimes K$. This criterion is satisfied by many groups.
