Csdrhrt

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

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

Many thanks in advance.

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 $Ain mathrm{GL}(2n,mathbb{C})$ satisfies $Amathrm{Sp}(n)A^{-1}subset mathrm{Sp}(n)$, 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 $hinmathrm{Sp}(n)$, 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 $hin mathrm{Sp}(n)$ 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 $Ainmathrm{GL}(n,mathbb{C})$ such that $Amathrm{SU}(n)A^{-1}= mathrm{SU}(n)$. 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 $phi(g) = hbar gh^{-1}$ for some $hinmathrm{SU}(n)$. However, it's easy to see that, for $n>2$, there is no pair $(A,h)$ such that $AgA^{-1} = hbar gh^{-1}$ for all $ginmathrm{SU}(n)$, so we only need to deal with the case $AgA^{-1} = h gh^{-1}$ with $hinmathrm{SU}(n)$ and $Ain mathrm{GL}(n,mathbb{C})$. Again, we find that $h^{-1}Ainmathrm{GL}(n,mathbb{C})$ 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.