Consider ridge regression with an additional constraint requiring that $\hat{\mathbf y}$ has unit sum of squares (equivalently, unit variance); if needed, one can assume that $\mathbf y$ has unit sum of squares as well:

$$\hat{\boldsymbol\beta}_\lambda^* = \arg\min\Big\{\|\mathbf y - \mathbf X \boldsymbol \beta\|^2+\lambda\|\boldsymbol\beta\|^2\Big\} \:\:\text{s.t.}\:\: \|\mathbf X \boldsymbol\beta\|^2=1.$$

What is the limit of $\hat{\boldsymbol\beta}_\lambda^*$ when $\lambda\to\infty$?

Here are some statements that I believe are true:

1. When $\lambda=0$, there is a neat explicit solution: take OLS estimator $\hat{\boldsymbol\beta}_0=(\mathbf X^\top \mathbf X)^{-1}\mathbf X^\top \mathbf y$ and normalize it to satisfy the constraint (one can see this by adding a Lagrange multiplier and differentiating): $$\hat{\boldsymbol\beta}_0^* = \hat{\boldsymbol\beta}_0 \big/ \|\mathbf X\hat{\boldsymbol\beta}_0\|.$$

2. In general, the solution is $$\hat{\boldsymbol\beta}_\lambda^*=\big((1+\mu)\mathbf X^\top \mathbf X + \lambda \mathbf I\big)^{-1}\mathbf X^\top \mathbf y\:\:\text{with $\mu$ needed to satisfy the constraint}.$$I don't see a closed form solution when $\lambda >0$. It seems that the solution is equivalent to the usual RR estimator with some $\lambda^*$ normalized to satisfy the constraint, but I don't see a closed formula for $\lambda^*$.

3. When $\lambda\to \infty$, the usual RR estimator $$\hat{\boldsymbol\beta}_\lambda=(\mathbf X^\top \mathbf X + \lambda \mathbf I)^{-1}\mathbf X^\top \mathbf y$$ obviously converges to zero, but its direction $\hat{\boldsymbol\beta}_\lambda \big/ \|\hat{\boldsymbol\beta}_\lambda\|$ converges to the direction of $\mathbf X^\top \mathbf y$, a.k.a. the first partial least squares (PLS) component.

Statements (2) and (3) together make me think that perhaps $\hat{\boldsymbol\beta}_\lambda^*$ also converges to the appropriately normalized $\mathbf X^\top \mathbf y$, but I am not sure if this is correct and I have not managed to convince myself either way.