Let us write $|A| = |B|$ iff there is a bijection $f \colon A \rightarrow B$. Working in $\operatorname{ZF + \neg C}$, can we prove that for any $A$ there is a transitive $B$ with $|A| = |B|$?
This is trivially true in $\operatorname{ZFC}$, since there is a bijection from $A$ to an ordinal. And in $ZF$, we may apply the Mostowski collapse to obtain such a bijection whenever there is an extensional and transitive relation on $A$. However, I have a feeling that my question might have a negative answer and if so, I'd like to learn a little about its consistency strength (relative to $\operatorname{ZF}$).
No. For example, an infinite transitive set can be mapped onto $\omega$. So amorphous sets, or other examples of Dedekind finite sets which cannot be mapped onto $\omega$ give you examples of sets which cannot be made transitive.
This can be pushed even further with ease. You could note that an infinite transitive set is in fact Dedekind infinite, thus extending the above example. And you can sit down and notice some of these things actually extend to ordinals larger than $\omega$ to encompass other strange sets.