No, there is a counterexample. The idea is that the use of the computation $X \le_T ran(g)$ can be much worse than identity, and since we only care about the reduction in one direction, we can drive that use up very large. In contrast, a function $f$ trying to maintain $X \equiv_T ran(f)$ can't freely increase the use on one side without making changes on the other.
We are building $g$ and $X$, and a basic module must diagonalize against a tuple $(f, \Phi, \Psi, k)$, ensuring that if $f$ is total and $g(y) < \min(f(y))$ for all $y > k$, then it's not the case that $\Phi(X) = \bigcup_n f(n)$ and $\Psi(\bigcup_n f(n)) = X$. For ease of notation, let $Z = \bigcup_n f(n)$.
For this module, we choose large $m_1 > m_0$ and enumerate the axiom $m_0 \not \in X$ with use $m_1 \not \in ran(g)$. We require that all future definitions of $g$ must use values greater than $m_1$, and we wait until we see $f$ converge on all $y \le k$ and all $y$ where we have already defined $g(y) < m_0$. We wait further until we see an expansionary stage: some $s$, $r_0$ and $r_1$ with $\Phi_s(X_s\upharpoonright r_0) = Z_s\upharpoonright r_1$ and $\Psi(Z_s\upharpoonright r_1) = X_s\upharpoonright m_0+1$. We pick an $m_2 > r_1$, we use our next definition of $g$ to enumerate $m_1$ into $ran(g)$, and we enumerate the axiom $m_0 \not \in X$ with use $m_2 \not \in ran(g)$. We then require that there be no enumerations into $X$ below $r_0$ and no further definitions of $g$ with values below $r_1$.
Now, we wait until $f$ converges on all $y$ where we have already defined $g(y) < r_1$. If $f$ uses any of these values to enumerate a new element into $Z$ below $r_1$, we then have $\Phi_s(X_s\upharpoonright r_0)$ incompatible with $Z$, so we win by maintaining the restraint on $X$ below $r_0$.
If $f$ does not take this opportunity to enumerate new elements into $Z$ below $r_1$, then $f$ will have lost the opportunity to ever again enumerate further elements into $Z$ below $r_1$ (assuming that our choice of $k$ was correct and we maintain the restraint on $g$). So $Z\upharpoonright r_1 = Z_s\upharpoonright r_1$. Now we enumerate $m_0$ into $X$ and $m_2$ into $ran(g)$, giving us $\Phi(Z)(m_0) = \Phi_s(Z_s\upharpoonright r_1)(m_0) = 0 \neq X(m_0)$, so we have won.
Now just arrange these modules into a finite injury priority argument.