I am reading Section 3.6 of the Naive Lie Theory by Stillwell. The author claims that we can argue that $T^n$ is the maximal torus of $Sp(n)$ exactly as in the case of $U(n)$. The proof of the maximal torus of $U(n)$ makes use of a matrix A in the maximal torus commuting with all elements R in the torus $T^{n}$ of $U(n)$, where R has the form $$R=\begin{pmatrix} e^{i\theta_1} & & & \\ & e^{i\theta_2} & & \\ & & \ddots & \\ & & & e^{i\theta_n} \\ \end{pmatrix}. $$ If A also has the form of R, then $T^{n}$ is the maximal torus of $U(n)$. After several steps, we arrive at the condition that $A\mathbf{e}_k=c_k\mathbf{e}_k$. Therefore $c_k=e^{i\phi_k}$ and we arrive at the conclusion.
However, in the case of $Sp(n)$, after carrying out the same steps, I arrived at the condition that $c_k=e^{\mathbf{u}_k\phi_k}$, where $\mathbf{u}_k$ is an arbitrary unit quaternion, and $A$ does not necessarily follow the form of $R$. I wonder how to argue that $\mathbf{u}_k=\mathbf{i}$ for all $k$'s.
Let $D$ be diagonal in $\mathrm{Sp}(n)$ with entries $\lambda_1,\cdots,\lambda_n$ which we may freely choose from $S^1$.
Suppose $S$ is an $n\times n$ quaternionic matrix which commutes with all such $D$. That is, $DS=SD$. This means that $\lambda_is_{ij}=s_{ij}\lambda_j$. If $s_{ij}\ne0$ for distinct $i,j$ then we could solve for $\lambda_j$ in terms of $\lambda_i$, which is impossible since we can choose the $\lambda$s independently of each other. Thus, $S$ is a diagonal matrix. Moreover, if we set $\lambda_i=\mathbf{i}$, then the equation $\mathbf{i}s_{ii}=s_{ii}\mathbf{i}$ implies $s_{ii}$ has zero $\mathbf{j}$ and $\mathbf{k}$ components, i.e. $S$ is a complex matrix. Then to finish, notice $s_{ii}\in S^1$ follows if $S\in\mathrm{Sp}(n)$.