I would be very grateful for a reference
to the following results (which are, I think, true,
though I never saw it in the literature).


Let $G\subset GL(n,{\Bbb C})$ be $U(n)$,
abd $A\in GL(2n,{\Bbb R})$ an endomorphism which satisfies
$AGA^{-1}=G$. Then $A\in {\Bbb R}^\* \times U(n)$.


Let $G\subset GL(n,{\Bbb C})$ be the group
$Sp(n)$ of quaternionic Hermitian matrices,
and $A\in GL(2n,{\Bbb R})$ an endomorphism which satisfies
$AGA^{-1}=G$. Then $A\in {\Bbb R}^\* \times Sp(n)\times Sp(1)$.


Many thanks in advance.



Stabilizer of Sp(n) and U(n) in GL(n)
First, let me fix a misunderstanding: $\mathrm{Sp}(n)$ does not sit in $\mathrm{GL}(n,\mathbb{C})$, but in $\mathrm{GL}(2n,\mathbb{C})$, so I'll assume that you mean, for the second part that $A$ lies in $\mathrm{GL}(2n,\mathbb{C})$.


These both follow immediately from the facts that all the automorphisms of 
$\mathrm{SU}(n)$ are either inner or conjugate-inner while the automorphisms of $\mathrm{Sp}(n)$ are are all inner.


It's a bit easier to deal with the $\mathrm{Sp}(n)$ case first since it has no outer automorphisms: If $A\in \mathrm{GL}(2n,\mathbb{C})$ satisfies $A\mathrm{Sp}(n)A^{-1}\subset \mathrm{Sp}(n)$, then consider the homomorphism $\phi:\mathrm{Sp}(n)\to \mathrm{Sp}(n)$ defined by $\phi(g) = AgA^{-1}$. This is a smooth, injective homomorphism, so it must be a smooth automorphism. Since every automorphism of $\mathrm{Sp}(n)$ is of the form $\phi(g) = hgh^{-1}$ for some $h\in\mathrm{Sp}(n)$, it follows that $AgA^{-1} = hgh^{-1}$, so $h^{-1}A$ lies in the commuting ring of $\mathrm{Sp}(n)$ (intersected with the $\mathbb{C}$-linear isomorphisms of $\mathbb{C}^{2n}$), and this is simply the nonzero complex multiples of the identity (since $\mathrm{Sp}(n)$ acts irreducibly on $\mathbb{C}^{2n}$). Thus, $A = \lambda h$ for some $h\in \mathrm{Sp}(n)$ and some nonzero complex scalar $\lambda$.


For the $\mathrm{U}(n)$ case, notice that the problem is equivalent to finding the conjugations that preserve the subgroup $\mathrm{SU}(n)$, so we might as well ask for the $A\in\mathrm{GL}(n,\mathbb{C})$ such that $A\mathrm{SU}(n)A^{-1}= \mathrm{SU}(n)$. Now, there is a slight complication, because for $n>2$, not all of the automorphisms of $\mathrm{SU}(n)$ are inner. For example, the automorphism $\psi(g) = \bar g$ is not inner. Instead, every automorphism is either of the form $\phi(g) = hgh^{-1}$ or $\phi(g) = h\bar gh^{-1}$ for some $h\in\mathrm{SU}(n)$. However, it's easy to see that, for $n>2$, there is no pair $(A,h)$ such that $AgA^{-1} = h\bar gh^{-1}$ for all $g\in\mathrm{SU}(n)$, so we only need to deal with the case $AgA^{-1} = h gh^{-1}$ with $h\in\mathrm{SU}(n)$ and $A\in \mathrm{GL}(n,\mathbb{C})$. Again, we find that $h^{-1}A\in\mathrm{GL}(n,\mathbb{C})$ must commute with all of the elements of $\mathrm{SU}(n)$, and this can happen only if $h^{-1}A$ is a multiple of the identity (again because of the irreducibility of the action of $\mathrm{SU}(n)$ on $\mathbb{C}^n$).


I think that the reason you haven't seen it in the literature is that it is such a direct consequence of well-known facts about Lie groups and representations.


