To be more specific, let $T$ be a first order theory and let $A$ and $B$ be two different first-order sentences, both in the same language as $T$ but independent of $T$. Additionally, suppose we have (working in some meta-theory) that Con($T+A$) implies Con($T+B$) (but not vice-versa).
$\textbf{Question}$: What, if anything, can we say about the sentences $A$ and $B$?
That is, can we only speak of the difference between $A$ and $B$ in terms of $T$? Are we limited to saying that adding $A$ to $T$ results in a theory which is "more likely" to derive a contradiction than adding $B$ to $T$? Or can we infer anything about $A$ and $B$ on their own? (Does one prove or refute the other? Might it be that $A$ has more first-order consequences than $B$ does?)
$\textbf{Follow-up question}$: Could it ever be the case that replacing $T$ with a different theory $S$ (also in the same language and unable to decide $A$ or $B$) leads to Con($S+B$) being equivalent to, or stronger than, Con($S+A$)?
Here is an example to better illustrate my questions:
It is known that the consistency strength of ZF together with the Axiom of Choice is exactly that of ZF:
$\text{Con(ZF+AC)} \iff \text{Con(ZF)}$
However, the consitency strength of ZF together with the Axiom of Determinacy is very much greater:
$\text{Con(ZF+AD)} \iff \text{Con(ZF}+\ \psi )$
where $\psi$ is the statement "there are infinitely many Woodin cardinals".
Do these relations of consitency strength tell us about AC and AD as sentences set apart from ZF? Is AD in some regard more powerful than AC (even though it's refutable in ZFC)? Could, for some new theory of sets $T$ in the same language as ZF, it be possible that $T$+AC is equiconsistent with $T$+AD?
The answers to these questions may turn out to be trivial, but I'm unable to settle on a satisfactory conclusion myself. Any responses offering a greater intuition about consistency strength would be very much appreciated.