In fact, except if $Ain Sp(n)$ 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, $Ax=xlambda$,
$xinmathbb{S}^{4n-1}$, then for any $muinmathbb{S}^3$, $Axmu=xmucdotoverline{mu}lambdamu$, so all the conjugates $overline{mu}lambdamu$ are also eigenvalues.
To see if there are eigenvalues at all, one can consider critical values of the function $xmapsto mathrm{Re}(x^*Ax)$ from $mathbb{S}^{4n-1}$ to $mathbb{R}$. Let $tin[-1,1]$ 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 $e^{itheta}$, 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].
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