This is a useful matrix inverse equality. $ \newcommand{\T}{\mathsf{T}} \newcommand{\R}{\mathbb{R}} $

Let $X \in \R^{n \times d}$ and let $\lambda > 0$. We have that $$ \begin{align} (X^{\T} X + \lambda I_d)^{-1} X^\T = X^\T (XX^\T + \lambda I_n)^{-1} \:. \label{eq:matrix_identity} \end{align} $$ This is simple to prove, and to do it we will use the SVD of $X$. Write $X = U \Sigma V^\T$, where $U \in \R^{n \times n}$, $\Sigma \in \R^{n \times d}$ and $V \in \R^{d \times d}$. Here, $U,V$ are orthogonal matrices, and $\Sigma_{ii}$ for $i=1, ..., \min(n, d)$ contains the $i$-th singular value of $X$, which we write as $\sigma_i \geq 0$. We also let $I_p \in \R^{p \times p}$ denote the $p \times p$ identity matrix for any $p \geq 1$. Then the LHS is simply $$ \begin{align*} (X^{\T} X + \lambda I_d)^{-1} X^\T &= ( V \Sigma^\T U^\T U \Sigma V^\T + \lambda I_d)^{-1} V \Sigma^\T U^\T \\ &= ( V (\Sigma^\T \Sigma + \lambda I_d) V^\T)^{-1} V \Sigma^\T U^\T \\ &= V (\Sigma^\T \Sigma + \lambda I_d)^{-1} V^\T V \Sigma^\T U^\T \\ &= V (\Sigma^\T \Sigma + \lambda I_d)^{-1} \Sigma^\T U^\T \:. \end{align*} $$ Similarly, the RHS is $$ \begin{align*} X^\T (XX^\T + \lambda I_n)^{-1} &= V \Sigma^\T U^\T (U \Sigma V^\T V \Sigma^\T U^\T + \lambda I_n)^{-1} \\ &= V \Sigma^\T (\Sigma\Sigma^\T + \lambda I_n)^{-1} U^\T \:. \end{align*} $$ Hence, to finish the proof it suffices to show that $$ \begin{align*} (\Sigma^\T \Sigma + \lambda I_d)^{-1} \Sigma^\T = \Sigma^\T (\Sigma\Sigma^\T + \lambda I_n)^{-1} \:. \end{align*} $$ But this is immediate. Both matrices are equal to a rectangular matrix $\Lambda \in \R^{d \times n}$ with entries $\Lambda_{ii} = \frac{\sigma_i}{\sigma_i^2 + \lambda}$ for $i=1, ..., \min(d, n)$ and zero everywhere else.

Edit: Originally I had claimed that $X(X^\T X + \lambda I_d)^{-1} X^\T = XX^\T (XX^\T + \lambda I_n)^{-1}$. While this is true, Daniel Seita pointed out to me that the stronger identity $\eqref{eq:matrix_identity}$ holds (e.g. pre-multiplication by $X$ is un-necessary).