Left and right eigenvalues A quaternionic matrix $A$ gives rise to a 
function $\mathbb{H}^n \to \mathbb{H}^n$
given by $x \mapsto A \cdot x$.  This is real linear, 
but not complex- or quaternionic-linear
(in general) if we consider $\mathbb{H}^n$ as 
a left $\mathbb{C}$ or $\mathbb{H}$ module, 
but is pretty good if we use right actions.
A right eigvenvalue of $A$ is a 
quaternion $q$ such that $A\cdot x = x \cdot q$
for some $x\in \mathbb{H}^n$; 
a left eigenvalue is quaterion $q$ 
such that $A \cdot x = q\cdot x$ for some  $x\in \mathbb{H}^n$.
The algebra of right eigenvalues is pretty good, 
but the algebra of left eigenvalues is quite
interesting.  For example, it is not hard to 
see that there are matrices $A$ with infinitely 
many left eigenvalues, even for $2$-by-$2$ matrices.
Now let's assume that $A\in Sp(n)$, so that the left eigenvalues are all contained in 
$S^3\subseteq \mathbb{H}$.  What sort of geometric properties must the set $L(A)$ of left
eigenvalues have?   
EDIT:  An example is 
$$
\begin{pmatrix} 
0 & 1 
\cr
-1 & 0
\end{pmatrix}   \cdot \begin{pmatrix} 
1 
\cr
q  
\end{pmatrix}
   = 
q \cdot
{\begin{pmatrix}
1
\cr
q  
\end{pmatrix}}$$
for any $q\in S^3 \subseteq \mathbb{H}$ with zero real part, since then $q^2 = -1$.
EDIT 2:  Examples like this show that for some symplectic matrices, the set of left 
eigenvalues is a union of copies of $S^2$.
 A: In fact, except if $A\in Sp(n)$ is an involution, there are also infinitely many right eigenvalues (all in $\mathbb{S}^3$). I realize that the question is about left eigenvalues, but as the following is too long for a comment, I post it as answer. Apologies.
If $\lambda$ is such an eigenvalue, $Ax=x\lambda$,
$x\in\mathbb{S}^{4n-1}$, then for any $\mu\in\mathbb{S}^3$, $Ax\mu=x\mu\cdot\overline{\mu}\lambda\mu$, so all the conjugates $\overline{\mu}\lambda\mu$ are also eigenvalues.
To see if there are eigenvalues at all, one can consider critical values of the function $x\mapsto \mathrm{Re}(x^*Ax)$ from $\mathbb{S}^{4n-1}$ to $\mathbb{R}$. Let $t\in[-1,1]$ be such a value, taken at the vector $x$. Then if $t=\pm 1$, $Ax=\pm x$, and if $|t|<1$,
$x$ and $Ax$ are independent over $\mathbb{R}$, and it is easy to see that the real plane they generate is invariant under $A$ (in fact, $(A+A^{-1})x=2tx$). Letting $t=\cos(\theta)$,
you can then find orthogonal $v_1,v_2$ of norm one in this plane such that $A$ has the standard rotation matrix $R_\theta$ in this basis. Then $v=v_1-v_2i$ is a right eigenvector with eigenvalue $e^{i\theta}$, and one can replace $i$ with any norm one imaginary quaternion. 
As in the unitary case, the (quaternionic) orthogonal hyperplane to $v$ is invariant, and you can induct on dimension to find a complete basis.
In short : conjugacy classes in $Sp(n)$ are parameterized by $n$ real numbers in [-1,1].
A: The symplectic 2x2 quaternionic matrices with infinite left eigenvalues are completely characterized [Macias-Pereira, Elect. J. Lin. Alg 2009]. There is a fundamental difference between left and right eigenvalues: the associate eigenvectors form a vector space (left eigenvalues) or not (right eigenvalues). In the latter case the quaternions similar to a rigth eigenvalue are eigenvalues too, but I should not call that to have infinite eigenvalues.
