Prove or disprove $SO(4n) \supseteq \frac{(Sp(1)\times Sp(n))}{\mathbb{Z}_2}$? I suspect that this is true
$$
\boxed{SO(4n) \supseteq \frac{(Sp(1)\times Sp(n))}{\mathbb{Z}_2}.}
$$
How do we prove it?

*

*When $n=1$, we have
$$
SO(4) \supseteq \frac{(SU(2)\times Sp(1))}{\mathbb{Z}_2}=  \frac{(SU(2)\times SU(2))}{\mathbb{Z}_2}=\frac{Spin(3) \times Spin(3)}{\mathbb{Z}_2}
=\frac{Spin(4)}{\mathbb{Z}_2}.
$$
which is true by isomorphism.


*When $n=2$, we have
$$
SO(8) \supseteq \frac{(SU(2)\times Sp(2))}{\mathbb{Z}_2}=  \frac{(SU(2)\times Spin(5))}{\mathbb{Z}_2}=\frac{Spin(3) \times Spin(5)}{\mathbb{Z}_2}.
$$
But in contrast only
$$
{Spin(8)} \supseteq \frac{Spin(3) \times Spin(5)}{\mathbb{Z}_2},
$$
which is true.
How to prove that general $n$, it is all true?
It is better even to show the explicit embedding.
p.s. Here I use the symplectic group $Sp(1)=SU(2)$ and $Sp(2)=Spin(5)$. We can see that a generic $SU(2)$ group can be represented by a rank-2 unitary matrix satisfies
$$V^\dagger V =\mathbb{I}.$$ Then we can write such a complex
$V = \begin{pmatrix} a &  b\\ - b^* & a~* 
\end{pmatrix}$. It can be checked that it obeys $Sp(1)$ condition
$$
V^T \begin{pmatrix} 0&  1\\ -1 & 0 \end{pmatrix} V = \begin{pmatrix} 0&  1\\ -1 & 0 \end{pmatrix}.
$$
 A: View $\mathbb{H}^n$ as a space of row vectors. This has a norm and sesqilinear inner product
$$ \|u\|^2=\langle u,u\rangle, \qquad \langle u,v\rangle=\overline{u_1}v_1+\cdots+\overline{u_n}v_n. $$
Note $\mathbb{H}^n$ is a left $\mathbb{H}$-vector space, and this form is conjugate-linear in the first argument. An $n\times n$ symplectic matrix $A\in\mathrm{Sp}(n)$ has the property that $\langle uA,vA\rangle=\langle u,v\rangle$ identically, which is equivalent to all the rows (and/or columns) of $A$ being orthonormal with respect to this inner product.
Note $\mathrm{Re}\langle u,v\rangle$ makes $\mathbb{H}^n$ a real inner product space with real dimension $4n$, and $T$ preserves this real inner product, so we have a homomorphism $S^3\times\mathrm{Sp}(n)\to\mathrm{O}(4n)$. Since the domain is connected, the range is in the identity component, so the homomorphism actually lands in $\mathrm{SO}(4n)$.
Suppose $(\lambda,A)$ is in the kernel, or in other words $\lambda vA=v$ for all $v\in\mathbb{H}^n$. Picking $v$ to be standard basis row vectors, we see the rows of $A$ are just $\lambda^{-1}$ times the standard basis vectors, in other words $A=\overline{\lambda}I_n$. Then $T(v)=\lambda v\overline{\lambda}$. Setting $v=\mu e_i$ (doesn't matter which $i$), this yields $\lambda\mu\overline{\lambda}=\mu$ for all scalars $\mu\in\mathbb{H}$. If $\lambda=\exp(\theta\mathbf{u})$ (where $\mathbf{u}\in\mathbb{R}^3$ is a pure imaginary quaternion), then $\lambda\mu\overline{\lambda}$ rotates $\mu$'s imaginary part by $2\theta$ around $\mathbf{u}$; the only way this has no effect is if $2\theta=0$ or $2\theta=2\pi$. In the latter case, $\lambda=-1$. Thus, the kernel of the homomorphism is $\pm(1,I_n)$, so by the first isomorphism theorem we can say
$$ (S^3\times\mathrm{Sp}(n))/\mathbb{Z}_2 \hookrightarrow \mathrm{SO}(4n). $$
I encourage you to write down the formula for this embedding with $n=2$.
