Suppose $f_i(x)$ is convex in $x$ for all $i\in\{1,\ldots,n\}$.
Consider a problem $$ \min_x\max_{i\in\{1,\ldots,n\}}f_i(x). $$ Indeed, this can be converted to a single minimization problem by introducing a variable $y\geq f_i(x),\ \forall i\in\{1,\ldots,n\}$, but is this equivalent to
$$ \max_{i\in\{1,\ldots,n\}}\min_x f_i(x)? $$
No, take $g(x) = x^2$, $f_1(x) = g(x-1)$, $f_2(x) = g(x+1)$.
Then $\min_x \max (f_1(x),f_x(x)) = 1$ and this occurs at $x=0$, whereas $\max_k \min_x f_k(x) = 0$ and this occurs at $x=\pm 1$.