Your conjecture is true. We assume PD throughout. The proof I see requires citing a number of facts from inner model theory and descriptive set theory. First, it uses Woodin's theorem characterizing the reals of $M_{2n-1}$ as the set $Q_{2n+1}$ of points in Baire space that are $\Delta^1_{2n+1}$ definable from a countable ordinal. In symbols,
Theorem (Woodin). $\omega^\omega\cap M_{2n-1} = Q_{2n+1}$.
It uses a correctness theorem for the odd levels:
Theorem $M_{2n-1}$ is $\Pi^1_{2n}$-correct.
A set $A\subseteq \omega^\omega$ is $\Pi^1_{2n+1}$-bounded if $\Pi^1_{2n+1} = \exists^{A} \Pi^1_{2n+1}$. We need that $Q_{2n+1}$ is $\Pi^1_{2n+1}$-bounded. In fact, something stronger is true:
Theorem (Kechris-Martin-Solovay). $Q_{2n+1}$ is the largest $\Pi^1_{2n+1}$-bounded subset of Baire space.
Finally, we need a slight generalization of Moschovakis's "Spector-Gandy theorem for the odd levels" (Moschovakis, Descriptive Set Theory, 6.E7) that follows from the same proof:
Theorem (Moschovakis) For any $\Pi^1_{2n+1}$-bounded subset $A$ of Baire space containing $\Delta^1_{2n+1}\cap P(\omega^\omega)$, $\Pi^1_{2n+1} = \exists^A\Pi^1_{2n}$.
Given these facts, the calculation becomes a straightforward pointclass calculation.
By definition, $\Sigma^1_{2n+1} = \exists^{\omega^\omega}\Pi^1_{2n}$, so $(\Sigma^1_{2n+1})^{M_{2n-1}} = \exists^{\omega^\omega\cap M_{2n-1}}(\Pi^1_{2n})^{M_{2n-1}}$.
Woodin's theorem characterizing $\Pi^1_{2n+1}$ along with the $\Pi^1_{2n}$-correctness of $M_{2n-1}$ imply $\exists^{\omega^\omega\cap M_{2n-1}}(\Pi^1_{2n})^{M_{2n-1}} = \exists^{Q_{2n+1}}\Pi^1_{2n}$.
The set $Q_{2n+1}$ contains $\Delta^1_{2n+1}\cap P(\omega)$ by definition and is $\Pi^1_{2n+1}$-bounded by the Kechris-Martin-Solvay theorem, so Moschovakis's Spector-Gandy Theorem implies that $\exists^{Q_{2n+1}}\Pi^1_{2n} = \Pi^1_{2n+1}$.
Stringing together a bunch of pointclass identities, one can conclude that $(\Sigma^1_{2n+1})^{M_{2n-1}} = \Pi^1_{2n+1}$.
You might also want to look at Theorem 4.12 of John Steel's paper Projectively Well-Ordered Inner Models. I think you can use the proof to get a more inner model theoretic proof of your conjecture, but Steel's result is closely related and of independent interest.