= Solution
The premise contains a wording error: $\lambda$ must be a nonzero \b[<eigenvalue>], rather than a nonzero <eigenvector>. Let $v\ne0$ satisfy
$$
X^TXv=\lambda v,\qquad \lambda\ne0.
$$
Then $Xv\ne0$, since otherwise $X^TXv=0$ would force $\lambda v=0$. Multiplying by $X$ gives
$$
XX^T(Xv)=X(X^TXv)=\lambda(Xv).
$$
Thus \b[$\boxed{\lambda\text{ is an eigenvalue of }XX^T}$], with <eigenvector> $Xv$. Conversely, applying $X^T$ maps an <eigenvector> of $XX^T$ with nonzero <eigenvalue> to one of $X^TX$.
These maps are inverses up to the factor $\lambda$ on the corresponding <eigenspaces>, so the nonzero <eigenvalues> have equal multiplicities. Their zero <eigenvalues> need not have the same multiplicity because the two <matrices> can have different sizes. For a unit <eigenvector> $v$ and real $X$, the normalized partner is $u=Xv/\sqrt\lambda$, since $\|Xv\|^2=v^TX^TXv=\lambda>0$.
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