Suppose $M_1^\#$ exists and is $\omega_1$-iterable.
Is it consistent that we can go to a generic extension $V[G]$ where $M_1^\#$ is no longer $\omega_1$-iterable?
Or "worse" $M_1^\#$ is no longer 2-iterable in Neeman's sense?
I suspect the answer is yes and that it will be relatively obvious. So more generally, I'm looking for places where problems like these might be addressed.
Andreas Lietz has already pointed out that $\omega_1$-iterability can fail in a generic extension. One can also consistently get $(\omega+1)$-iterability to fail:
Suppose there is a transitive model of ZFC + "$M_1^\#$ exists and is $(0,\omega_1)$-iterable", and let $M$ be such with minimal ordinal height. It is easy to see we may assume that $M=L_\alpha[X]$ for some set $X\in M$ of ordinals, where $\alpha=\mathrm{OR}^M$, and we can take $X\subseteq\lambda$ for some strong limit cardinal $\lambda$ of $M$, and $X$ coding $V_\lambda^M$. Let $N=(M_1^{\#})^M$. Let $G$ be $(M,\mathrm{Coll}(\omega,\lambda))$-generic.
I claim that in $M[G]$, $N$ is not $(\omega+1)$-iterable with respect to trees based on $N|\delta^N$ (the Woodin of $N$). For suppose otherwise. We have $M[G]=L_\alpha[x]$ for a certain real $x$, and we can take $x$ to code $X$ in a simple manner. So since $\mathcal{P}(\delta^N)\cap N$ is countable in $M[G]$, working in $M[G]=L_\alpha[x]$, we can form a Neeman genericity iteration on $N$ (see Theorem 3.3 of "AN INNER MODELS PROOF OF THE KECHRIS-MARTIN THEOREM"), producing a tree $\mathcal{T}$ on $N$ of length $\omega$, such that if $b$ is any $\mathcal{T}$-cofinal wellfounded branch, then there is a $\mathrm{Coll}(\omega,\delta^{M^{\mathcal{T}}_b})$-generic $g\in L_\alpha[x]$ with $x\in M^{\mathcal{T}}_b[g]$. But letting $\kappa$ be the critical point of the active extender of $M^{\mathcal{T}}_b$, then note that $\kappa<\alpha$ and $L_\kappa[x]\models$ ZFC and $X\in L_\kappa[x]$. Therefore $L_\kappa[X]\models$ ZFC. But since $\kappa<\alpha$, we have $L_\kappa[X]\subseteq M$, so $V_\lambda^{L_\kappa[X]}=V_\lambda^{M}$, so $L_\kappa[X]\models$ ZFC + "$M_1^{\#}$ exists and is $(0,\omega_1)$-iterable", contradicting the minimality of $\alpha$.
However, one should note that if $M\models$ ZFC + "$M_1^{\#}$ exists and is $(0,\omega_1)$-iterable" and there is a larger model $W$ of ZFC with $M\subseteq W$ and $\mathrm{OR}^M=\mathrm{OR}^W$ and $(M_1^{\#})^M$ is fully iterable in $W$ (so $(M_1^{\#})^M=(M_1^{\#})^W$) then for any set generic extension $M[G]$ with $\mathbb{R}^{M[G]}\subseteq W$, we will have that $(M_1^{\#})^M$ is $(0,\omega+1)$-iterable in $M[G]$. This is just by $\Sigma^1_2$ absoluteness between $M[G]$ and $W$: given a length $\omega$ tree $\mathcal{T}\in M[G]$, there is a $\mathcal{T}$-cofinal wellfounded branch $b\in W$, but this is just a $\Sigma^1_2$ assertion about $\mathcal{T}$ which is true in $W$, hence true in $M[G]$. Of course, the "sharp" at the top of $M^{\mathcal{T}}_b$ need not be iterable.
Yes, this is possible. If $M_1^\#$ exists and is $\omega_1$-iterable then $x^\#$ exists for all reals $x$: By genericity iterations (either Woodin's or Neeman's version), for any real $x$ there is a countable iterate $N$ of $M_1^\#$ so that $x$ is generic over $N$ for a forcing of size the Woodin cardinal of $N$. So $N[x]$ still "has a sharp" and $x^\#$ exists.
If now, e.g. $M_1^\#$ exists, is $\omega_1$-iterable and $V=L[A]$ for a set $A$, then in the extension by $\mathrm{Col}(\omega, A)$, $(M_1^\#)^V$ cannot be $\omega_1$-iterable.
