A question about the definition of equivariant map If $S$ is a set of functions from $X$ to $Y$ then I can consider the action of a group $G$ on $S$ via its action on $X$ and $Y$ by the formula
$$(g \cdot f)(x) = g \cdot f(g^{-1} \cdot x),$$
So we are considering left actions both on $X$ and $Y$.
Then, the definition of equivariant map pops out but I don't really understand how it is related to previous statement.
An function $f: X \rightarrow Y$ is equivariant if it satisfies
$$f(g \cdot x) = g \cdot f(x) \, \, \, \, \forall g \in G.$$
What's happening here? Are we assuming that the group $G$ acts trivially on $Y$? How to get this definition from the previous statement?
 A: So, recapitulating, we have actions of $G$ on $X$ and $Y$. This induces an action of $G$ on $S=\{f:X\longrightarrow Y\}$ by setting
$$(g\cdot f)(x)=g\cdot f(g^{-1}\cdot x)$$
On the other hand, a function $f\in S$ is said to be equivariant provided
$$f(g\cdot x)=g\cdot f(x) \quad \forall g\in G$$
Taking $x=g^{-1}\cdot x^{*}$ in this last expression yields
$$f(x^{*})=g\cdot f(g^{-1}\cdot x^{*})=(g\cdot f)(x^{*})$$
Thus, a function $f\in S$ will be equivariant if and only if $f=g\cdot f$ for all $g\in G$, i.e., if it is $G$-invariant under the above defined action of $G$ on $S$.
A: The two definitions are a priori independent, the only thing you can deduce is that for the induced action $(g,f)\mapsto g*f$ we don't have $(g*f)(x)=f(g\cdot x)$ in general.
Instead, by the first definition, we have
$$(g*f)(x)\ =\ g\cdot f(g^{-1}\cdot x)\,.$$
Note that the second definition doesn't say anything about the induced action of the first definition.
However, as pointed out in the comments, there's a connection between these definitions, namely

A map $f:X\to Y$ is equivariant iff
$$g* f=f$$
for every $g\in G$.

Indeed, if $f$ is equivariant, then
$$(g* f)(x)=g\cdot f(g^{-1}\cdot x)=f(g\cdot g^{-1}\cdot x)=f(x)\,.$$
And if $f$ is stabilized by $G$, then in particular for any $g^{-1}\in G$ we have $f=g^{-1}* f$, so
$$f(x)=(g^{-1}* f)(x)=g^{-1}\cdot f(g\cdot x)\implies g\cdot f(x)=f(g\cdot x)\,.$$
