Let $\kappa$ be a measurable cardinal and $j: V \rightarrow M$ be the elementary embedding with critical point $\kappa$. Let $X$ be a set such that for all $x\in X$, $j(x) = x$. My question is: when is $X$ a member of $M$?
We know that $X \in M$ when $|X| \leq \kappa$. But given everything in $X$ is unmoved by $j$, can $X$ be larger, or even arbitrarily large? In particular, is there such an $X$ of cardinality $\kappa^+$?
On
Suppose that $j$ is an ultrapower embedding given by $U$. We know that: (1) $U$ has size $2^\kappa$, and (2) $U\notin M$.
Consider the set $T$ of the first $2^\kappa$ limit cardinals which are also fixed points of $j$, and let $\lambda_\xi$ be the $\xi$th member of $T$. Next fix a set of ordinals $U'$ such that the transitive collapse of $U'$ is the transitive closure of $\{U\}$ (which happens to be $\{U\}\cup U\cup\kappa$).
Finally, define $X=T\cup\{\lambda_\xi^+\mid \xi\in U'\}$. First note that if $\lambda$ is a fixed point, then $\lambda^+$ is a fixed point as well, so for all $x\in X$, $j(x)=x$. But we can read $U'$ from $X$, since we can enumerate all the limit cardinals in $X$, and $U'$ will be the set of the indices of those that also have their successor in $X$.
But since from $U'$ we can read $U$, and $U\notin M$, we have that $X\notin M$ either.
This is robust, in the sense that even if $j$ is not an ultrapower embedding, we can choose any $U\in V\setminus M$ and code it in a similar way.
(There are probably simpler examples, though.)
In general, we can certainly have such a set in $M$, since if $j:V\to M$ is a $\lambda$-supercompactness embedding then in fact $M^\lambda\subseteq M$, so if $\kappa$ is supercompact then we can get such embeddings and sets $X$ with $X$ as large as desired.
You said "We know $X\in M$ when $|X|\leq\kappa$". But this isn't true in general. Maybe you are assuming that $M=\mathrm{Ult}(V,U)$ for some nonprincipal $\kappa$-complete ultrafilter $U$ over $\kappa$? Under this assumption, there can be no such $X\in M$ of cardinality $>\kappa$:
Theorem. Let $M=\mathrm{Ult}(V,U)$ where $U$ is a nonprincipal $\kappa$-complete ultrafilter over $\kappa$. Let $j:V\to M$ be the ultrapower embedding. Let $X\in M$ be a set such that $j(x)=x$ for all $x\in X$. Then $V\models$"$|X|\leq\kappa$" and hence $M\models$"$|X|\leq\kappa$" (since $M^\kappa\subseteq M$).
(The proof below uses an idea from the proof of Lemma 2.2/Theorem 2.3 of Reinhardt cardinals and iterates of V.)
Proof. Suppose otherwise. Then by considering a wellorder of $X$ in $M$ and cutting $X$ down to an initial segment of $X$ under this wellorder, we can assume $V\models$"$|X|=\kappa^+$". Now fix a bijection $\pi:\kappa^+\to X$ in $V$. Then $M\models$"$j(\pi):j(\kappa^+)\to j(X)$ is a bijection".
Claim: $j(\pi)^{-1}``X=j``\kappa^+$.
Proof of Claim: Equivalently, since $j(\pi)$ is a bijection, we must show $j(\pi)\circ j``\kappa^+=X$. For this, first observe that by the elementarity of $j$, we have $$j(\pi(\alpha))=j(\pi)(j(\alpha))$$ for all $\alpha<\kappa^+$, and therefore $$j\circ\pi=j(\pi)\circ j\upharpoonright\kappa^+.$$ But $j\circ\pi=\pi$, since $j\upharpoonright X=\mathrm{id}$, so $$\pi=j(\pi)\circ j\upharpoonright\kappa^+,$$ so $$X=\pi``\kappa^+=j(\pi)\circ j``\kappa^+,$$ as desired.
Now by the claim, and since $j(\pi),X\in M$, we get $j``\kappa^+\in M$, but this is false, because $j``\kappa^+$ is cofinal in $j(\kappa^+)$, which is a regular cardinal in $M$, and $j``\kappa^+$ has ordertype $\kappa^+$, and $\kappa^+<j(\kappa^+)$, so $j``\kappa^+$ singularizes $j(\kappa^+)$ in $M$, a contradiction.
(Remark: Actually if $M$ contains the element $j``X$ where $X$ has cardinality $\lambda$ in $V$, then $j``\lambda\in M$, and so $j\upharpoonright\lambda\in M$; this is by a slight variant of the argument above; note we don't need that $X$ is fixed pointwise for this.)