**not an answer to the revised question**

But Das Curious asks for the counterexample (to the **original question**) when the countability assumption is omitted. Here it is.

Let $X = [0,1)^{[0,1]}$, that is: the set of all functions $x \colon [0,1] \to [0,1)$. Let $\lambda$ be Lebesgue measure on $[0,1)$. Let $\Lambda$ be the product measure on $X$ with the product $\sigma$-algebra $\mathcal F$. In particular: if $A \in \mathcal F$, then there is a countable $J \subseteq [0,1]$ such that $A$ "depends only on" $J$ in the sense that: for $x,y\in X$ with $x(t)=y(t)$ for all $t \in J$, then either $x,y$ both belong to $A$ or both belong to $X \setminus A$.

Let $\theta$ be the "irrational rotation" $\theta \colon [0,1) \to [0,1)$ defined by $t \mapsto t+\sqrt{2} \pmod{1}$. So $\theta$ is a measurable, measure-preserving bijection (with no fixed point) of $([0,1),\lambda)$ onto itself. For a fixed $s \in [0,1]$, define $\theta_s \colon X \to X$ by "rotation of coordinate $s$", namely
$\theta_s(x)(s) = \theta(x(s))$ and $\theta_s(x)(t) = x(t)$ for all $t \ne s$. So $\theta_s$ is a measurable, measure-preserving bijection of $(X,\Lambda)$ onto itself.

For $s \in [0,1)$ let $Q_s = \{x \in X\;|\; x(1)=s\}$. Thus $(Q_s)_{s \in [0,1)}$ is a family of subsets of $X$; the family has cardinal of the continuum, so it is indexed by $[0,1)$; each $Q_s$ is a measurable set with $\Lambda(Q_s)=0$; the sets $Q_s$ are pairwise disjoint; but the union $\bigcup_{s} Q_s = X$ is the whole space.

Now we can define $T \colon X \to X$. Let
$$
T(x)(t) = \theta(x(t))\qquad\text{if } x \in Q_{t}
\\
T(x)(t) = x(t)\qquad\text{if }x \notin Q_t
$$
Now each $x$ belongs to exactly one $Q_t$, so $x$ and $T(x)$ agree, except on a single coordinate, where they disagree. Note $T(x) \ne x$ for all $x$. Because $1 \notin [0,1)$, the map $T$ maps each $Q_s$ into itself. Thus $T$ is a bijection of $X$ onto itself: the inverse map has the same definition using $\theta^{-1}$ in place of $\theta$.

We claim that the map $T$ is measurable. Let $A \in \mathcal F$. We claim $T^{-1}(A) \in \mathcal F$. There is a countable set $J \subseteq [0,1)$ such that $A$ depends only on $J$. For $s \in J$ let $A_s := A \cap Q_s$. Let $B := A \setminus \bigcup_{s \in J} Q_s$. Now $A$ is the countable union
$$
A = B \cup \bigcup_{s \in J} A_s
$$
where $B$ and all $A_s$ belong to $\mathcal F$.
The inverse image $T^{-1}(A)$ is the countable union
$$
T^{-1}(A) = T^{-1}(B) \cup \bigcup_{s \in J} T^{-1}(A_s) .
$$
Now by the definition of $J$, we have $T^{-1}(B)=B$. Each $\theta_s$ is a measurable function, and
$T^{-1}(A_s) = \theta_s^{-1}(A_s)$, so each $T^{-1}(A_s) \in \cal F$. Therefore $T^{-1}(A)$ is written as a countable union of sets from $\mathcal F$.

We claim $A = T(A)$ up to null sets for any $A \in \mathcal F$.
Let $J$ be as before.
If $x$ is not in the null set $\bigcup_{s \in J} Q_s$, then either $x,T(x)$ are both in $A$ or both in $X \setminus A$. So $A$ and $T(A)$ agree except for adding and subtracting parts of that null set.