Let $f:\mathrm{R}^n \rightarrow \mathrm{R}$, be convex and $g:\mathrm{R}^n \rightarrow \mathrm{R}^m$ with $n>m$. Denote $x^*$ as the solution to
$$\min_x f(x) \qquad \text{s.t. } g(x)=0.$$
Introduce new variables $x=y+z$. Does, for all $y^*$, $z^*$ being the solution of
$$ \min_{y,z} f(y+z) \qquad \text{s.t. } g(y+z)=0 $$
hold, that $y^*+z^*$ is also a solution of the first optimization problem? The other way round seems trivial, for all $x^*$ fulfilling the first optimization problem, we can choose any $y^*$, $z^*$ fulfilling $x^*=y^*+z^*$ that minimize the second optimization problem.
Yes, both problems are equivalent, since
\begin{align} &\min\{ f(y+z) \colon y,z \in \mathbb{R}^n, ~ g(y+z) = 0 \} \\ &= \min\{ f(\tilde{x}) \colon \tilde{x} = y+z, ~ y,z \in \mathbb{R}^n, ~ g(\tilde{x}) = 0\} \\ &= \min\{ f(x) \colon x \in \mathbb{R}^n, ~ g(x) = 0\} \end{align}
and you can always find $y,z \in \mathbb{R}^n$ such that $\tilde{x} = y+z$.