[Reposted from math.stackexchange]

Consider a monoid $M$ acting on a set $X$, where $M$ is the full transformation monoid on some set $A$.

Say that $B\subseteq A$ fixes $x\in X$ iff, for all $m\in M$, if $m(b) = b$ for all $b\in B$, then $mx=x$.

Say that $B\subseteq A$ pins down $x\in X$ iff, for all $m,n\in M$, if $m(b)=n(b)$ for all $b\in B$, then $mx=nx$.

[Apologies if there's more standard terminology for these notions]

Question 1: If $B$ fixes $x$, does it follow that $B$ pins down $x$?

Question 2: If $B$ and $B'$ both pin down $x$, does it follow that $B\cap B'$ pins down $x$?

[Reposted from math.stackexchange]

Consider a monoid $M$ acting on a set $X$, where $M$ is the full transformation monoid on some set $A$ (i.e., the set of all functions from $A$ to itself, with function composition as the monoid operation).

Say that $B\subseteq A$ fixes $x\in X$ iff, for all $m\in M$, if $m(b) = b$ for all $b\in B$, then $mx=x$.

Say that $B\subseteq A$ pins down $x\in X$ iff, for all $m,n\in M$, if $m(b)=n(b)$ for all $b\in B$, then $mx=nx$.

[Apologies if there's more standard terminology for these notions]

Question 1: If $B$ fixes $x$, does it follow that $B$ pins down $x$?

Question 2: If $B$ and $B'$ both pin down $x$, does it follow that $B\cap B'$ pins down $x$?