Is every true statement independent of $PA$ equivalent to some consistency statement? Most true statements independent of PA that I know of is equivalent to some consistency statement. For example


*

*Con(PA), Con(PA + Con(PA)), Con(PA + Con(PA) + Con (PA + Con(PA)), $\dots$

*Goodstein's theorem is equivalent to Con(PA)

*Any conjunction or disjunction of the above.


Is every true statement independent of PA equivalent to some consistency statement?
By "equivalent to some consistency statement", I mean that $PA \vdash S \iff Con(T)$, for some theory $T$. Also, $T$ should be either finite, or specified by a Turing machine that outputs its axioms (and such that PA proves that the Turing machine never stops outputting statements), so that the description of $T$ doesn't throw PA off.
EDIT: In particular, are there are $\Pi^0_1$ examples?
 A: The theory $PA + Con(PA)$ has the property you are asking for, this is the so called Friedman-Goldfarb-Harrington principle (see, e.g., Fifty years of self-reference in arithmetic, p. 366). Formally, for every $\Pi_1$ sentence $\pi$, there is a $\Pi_1$ sentence $\psi$ such that $PA + Con(PA) \vdash \pi \leftrightarrow Con(PA + \psi)$.
EDIT:
That being said, my hunch is that the answer must be "no" for plain $PA$ even though I can't produce a counterexample at the moment.
As observed by Will Sawin, we have that for every $\Pi_1$ sentence $\pi$, there is a $\Pi_1$ sentence $\psi$ such that $PA \vdash \pi \leftrightarrow Con(EA + \psi)$, which gives a positive answer to the OP's modified question.
My hunch is that $Con(EA + \psi)$ can not be replaced by $Con(PA + \psi)$ in the above.
A: $1$-consistency of $PA$ is a true $\Pi_3$ sentence which is not provable in $PA$+{all true $\Pi_1$ sentences} (see this article). Simple (iterated) consistency statements (as you mentioned above) are all (true) $\Pi_1$ sentences, so it is not equivalent to any $\Pi_1$ sentence. 
