Element of a ring acting as a permutation on an ideal I am investigating cases when $r \cdot I = I$ for some element $r$ and an ideal $I$ of a commutative ring or rng R.
Clearly, $r \cdot \langle 0 \rangle = \langle 0 \rangle$ for any element $r$ of $R$, and $u \cdot R = R$ for any unit $u$.  
Let's call elements that generate the same principal ideal associates.
$[r]$ is the equivalence class of associates of an element $r$.
Let $M$ be the sum of all ideals $I$ such that $r \cdot I = I$.
$M$ is the maximal ideal with respect to the property.
Obviously, if $r \cdot M = M$, then $[r] \cdot M = M$.
Every element of $[r]$ acts as a permutation on $M$.
Let's check some examples on cyclic rings:


*

*$R = \mathbb Z_{10}, [r] = [2] = \{2,4,6,8\}, M = 2 \mathbb Z_{10} = \{0,2,4,6,8\}$:
$2 \cdot M = (2\ 4\ 8\ 6)$
$4 \cdot M = (6\ 8\ 4\ 2)$
$6 \cdot M = Id$
$8 \cdot M = (2\ 8)(4\ 6)$

*$R = \mathbb Z_{10}, [r] = [5] = \{5\}, M = 5 \mathbb Z_{10} = \{0,5\}$:
$5 \cdot M = Id$ 

*$R = \mathbb Z_{12}, [r] = [2] = \{2,10\}, M = 4 \mathbb Z_{12} = \{0,4,8\}$:
$2 \cdot M = (4\ 8)$
$10 \cdot M = Id$ 

*$R = \mathbb Z_{12}, [r] = [3] = \{3,9\}, M = 3 \mathbb Z_{12} = \{0,3,6,9\}$:
$3 \cdot M = (3\ 9)$
$9 \cdot M = Id$ 
Questions:
1. Is the set of permutations $[r] \cdot M = M$ aways a group?
2. If $[r] \cdot M = M$, does it mean $[r] \cdot I = I$ for any ideal $I \subseteq M$?
3. If $\langle 0 \rangle \subsetneq M \subsetneq R$ for an element $r$, does it mean $r$ is irreducible?
4. Are there any interesting properties of such elements and ideals?
 A: First, some setup. $M$, as an ideal of $R$, is an $R$-module. Multiplication of $M$ by any element of $R$ (with your property or not) is an endomorphism of $M$ (as an abelian group, because $r(m_1+m_2) = rm_1+rm_2$), and because $R$ is commutative, it's even an $R$-module endomorphism (since $r(r'm) = r'(rm)$). Thus sending $r$ to its multiplication action on $M$ gives us a set map
$$\Phi: R\rightarrow \operatorname{End}_R(M).$$
Furthermore, right-distributivity and associativity in $R$ mean that $\Phi$ is a rng homomorphism, and if $R$ has a unit, it's even a ring homomorphism.
The statement that $r\cdot M = M$ is the statement that $\Phi(r)$ is surjective from $M$ to $M$.
Now, one thing that needs to be cleared up: If $r\cdot M = M$, it does not imply that $r$'s action on $M$ (which I have named $\Phi(r)$) is a permutation. It must be surjective, but it could fail to be injective. Example: let 
$$ R = \mathbb{Z}[x,y_1,y_2,y_3,\dots] / (xy_1, y_1-xy_2, y_2-xy_3, \dots),$$
and let $M$ be the ideal generated by the $y$'s, and take $r=x$. Then multiplication by $r=x$ is surjective onto $M$ since $y_i = xy_{i+1}\in x\cdot M$ for all $i$. However, $xy_1=0$, so $r=x$'s action on $M$ has a nontrivial kernel. 
So, in full generality, we need to separate the question about what is implied by the assumption that $r\cdot M = M$, versus the stronger assumption that $r$ acts as a permutation on $M$.
On the other hand, if $R$ is noetherian, then $r\cdot M=M$ does imply that $r$ acts as a permutation on $M$, which we can see as follows. If $r\cdot M = M$, then $\Phi(r)$ is a surjective $R$-module map from $M$ to itself. In this situation, the first isomorphism theorem tells us that $M \cong M/\ker \Phi(r)$ as $R$-modules. If $r$ acts non-injectively on $M$, i.e., if $\ker \Phi(r)$ is nontrivial, then this isomorphism means $r$ also acts non-injectively on $M/\ker\Phi(r)$, i.e., multiplication by $r$ puts something in $\ker\Phi(r)$ that wasn't already there. In other words, $\ker\Phi(r^2)$ is strictly bigger than $\ker\Phi(r)$. Similar reasoning shows that $\ker\Phi(r^{j+1})$ is strictly bigger than $\ker\Phi(r^j)$, for all $j$. Thus
$$\ker\Phi(r)\subset\ker\Phi(r^2)\subset \dots$$
is a nonterminating, strictly increasing chain of $R$-submodules of $M$. But an $R$-submodule of $M$ is an ideal of $R$, because $M\subset R$, so if $R$ is noetherian, this is impossible.
So, if you are willing to assume $R$ is noetherian, then we can go ahead and treat "$r\cdot M=M$" and "$r$ acts as a permutation on $M$" as equivalent statements.
For the remainder of this answer, I will assume the stronger version: that $r$ acts as a permutation on $M$, i.e., that $\Phi(r)$ is bijective. (Your interest is in $M$'s that are maximal with respect to this property.) However, I will not assume that $R$ is noetherian.
Finally, your questions:
1) For a given $r$ and $M$, does the set of permutations coming from multiplication of $r$'s associates by $M$ necessarily form a group? Answer: no. Let
$$ R = \mathbb{Q}[x;\dots,y_{-2},y_{-1},y_0,y_1,y_2,\dots]/(\dots,y_{-1} - xy_0, y_0 - xy_1, y_1-xy_2,\dots),$$
and let $M$ be the ideal generated by the $y_i$'s. Let $r=x$. Then $r\cdot M=M$ by similar logic as above: every $y_i$ is the image of multiplication of $y_{i+1}$ by $x$. Furthermore, $M$ is maximal with respect to this property: if $M'$ is any ideal properly containing $M$, then $M'$ has nonzero image in $R/M = \mathbb{Q}[x]$, and then this image is a nonzero ideal in $\mathbb{Q}[x]$ on which multiplication by $x$ acts surjectively. But this doesn't exist: any nonzero ideal in $\mathbb{Q}[x]$ contains a polynomial of minimal degree, and multiplication by $x$ raises the degree, so there is no nonzero ideal in $\mathbb{Q}[x]$ on which multiplication by $x$ acts surjectively. So this is an example of $r,M$ that fits your framework.
However, there is no associate of $r=x$, nor any element of $R$ at all, that inverts $x$'s action on $M$. Indeed, e.g., $y_0\in M$, and $xy_0=y_{-1}$, but there is no $r\in R$ such that $ry_{-1} = y_0$.
2) No. In the previous example, take $I$ to be the ideal generated by $y_0,y_{-1},\dots$. Then $r\cdot I$ does not contain $y_0$.
Here's even a noetherian example, which shows that $r\cdot I$ doesn't even have to be contained in $I$: let
$$ R = \mathbb{Q}[t,x,y]/(x-ty, y-tx).$$
Let $M$ be the ideal generated by $x$ and $y$, and let $r=t$. Then $r\cdot M = M$, but within $M$, $I=(x)$ and $J=(y)$ are transposed with each other by $r=t$, not fixed.
3) No. Take the $R$ of the previous (non-noetherian) example and consider $S = R[z,w]/(x-zw)$, and extend $M$ to $S$ (i.e., replace the previous $M$ with the ideal generated by the $y_i$'s in $S$ instead of $R$). (Retain $r=x$.) Then it is still true $M$ is maximal with respect to the property that $r\cdot M = M$, by similar logic: $S/M$ is $\mathbb{Q}[x,z,w]/(x-zw)\cong \mathbb{Q}[z,w]$. This is again a ring in which there is no ideal that is surjected onto itself by multiplication by $x=zw$. However, $r=x=zw$ is definitely not irreducible.
The same type of argument should also work for the noetherian example. Take $S = R[z,w]/(t-zw)$.
4) I don't have a comprehensive answer. (Who decides what's interesting anyway? :) One observation is that if $M$ is nonzero, but finitely generated as an $R$-module (for example, if $R$ is noetherian), then it is not possible for $r$ to lie in the Jacobson radical of $R$. (If $M$ is finitely generated and $rM = M$ with $r$ in the Jacobson radical, then $M$ would $=0$ by Nakayama's lemma.) Therefore, there is some maximal ideal $\mathfrak{m}$ with $r\notin \mathfrak{m}$.
