When will L2 norm and squared L2 norm be equivalent in ridge regression? I am doing ridge regression and am wondering why it uses squared L2 norm. This post Why l2 norm squared but l1 norm not squared? gave some thought about why using it. But a further question arises: when will using L2 and squared L2 norm be equivalent?
In my case, I am dealing with a completely orthonormal case. It is a cell-means model with the objective function in the form $\sum_i\sum_j(X_{ij}-\mu_i)^2$:

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*Using squared L2 norm adds a term $\lambda(\mu_1^2+\cdots+\mu_p^2) $ to the objective function, which then reduces to p separate univariate optimizations.

*Using L2 norm adds a term $\lambda\sqrt{(\mu_1^2+\cdots+\mu_p^2)} $, which cannot be reduced to univariate optimizations because of the square root of sum of all parameters.

In this case, I don't think those two are still equivalent. So which one to use?
(I think from the KKT condition we should conclude that those two are equivalent, so is there something in the orthonormal case that violates some assumptions?)

Assume we find an optimum $\lambda_{opt}$ for the ridge regression case (squared l2 norm) using out of sample error etc, and associated coefficients $\beta_{opt}$
The (gradient) optimality condition for the ridge coefficients is
$$\frac{\partial}{\partial \beta}SSE + 2\lambda_{opt}\beta=0$$
Similarly the gradient optimality condition for the l2 norm case for some regularisation partameter $\mu$ is
$$\frac{\partial}{\partial \beta}SSE + \mu\frac{\beta}{|\beta|_2}=0$$
so $\beta_{opt}$ is also optimal for the l2 norm case with $\mu = 2\lambda_{opt}|\beta_{opt}|_2$.
So every l2 norm squared case has a corresponding l2 norm solution. Reversing the argument, we see that the reverse is also true.
So if we find an optimal set of coefficients using ridge regression that minimise eg out of sample error, then we will find the same  optimal set of coefficients using l2 norm regression. (note the complication raised by @SextusEmpiricus wrt crossvalidation)
