Consider the following two optimization problems
$$\underset{x}{\text{min}} f(x) \quad \text{s.t.}\quad Ax=b$$
$$\underset{x}{\text{min}}\left[ f(x) + \mathbb{\lambda}^T(Ax-b)\right]$$
with $f(x)$ any continuous differentiable function of $x\in \mathbb{R}^n$, $b\in \mathbb{R}^m$, $A\in \mathbb{R}^{m\times n}$. $\lambda \in \mathbb{R}^m$ is a vector of weights that should be chosen to minimize the argument. Is it correct to say these two problems are equivalent because they both have the same Lagrangian?
$$L(x) = f(x) + \lambda^T(Ax-b)$$
UPDATE: I would like to revise the question, after receiving a few comments, to refer to these optimization problems instead
$$\underset{x}{\text{min}} f(x) \quad \text{s.t.}\quad Ax=b$$
$$\underset{x}{\text{min}}\left[ f(x) + ||Ax-b||_2\right]$$
where there exists $x$ such that $Ax=b$.
Generally speaking, they are NOT equivalent. Every optimal solution of the first problem ($x^*$) would satisfy $Ax^* = b$ and thus, would yield the same objective function value ($f(x^*)$) for both problems, but this may not be optimal for the second problem. Going the other way round, every optimal solution for the second problem may not always be an optimal solution for the first problem. Also note that the second problem has the $\lambda$s included in its decision variables. This is not the case for the first problem.
This argument is based on the assumption that two optimization problems are equivalent if their optimal value(s) and optimal solution(s) are the same. Things may be different under different equivalency criteria.
For example: $\min_x x^2$ s.t. $x = 2$ is minimum at $x=2$ and has minimum value $4$. However $\min_{x,\lambda} x^2 + \lambda(x-2)$ has minimum value tending to negative infinity when $x = -\lambda/2$ and $\lambda$ tends to positive infinity.