Action of a symmetric group in operad For a symmetric monoidal category $(V,\otimes, I)$,
we have a definition of symmetric operad as:


*

*collection of objects $P(n) \in  V$

*unit map $I \to P(1)$

*product map $P (n) \otimes P(j_1)\otimes ....P(j_n) \to P(j)$ where $j=j_1+\cdots +j_n$

*and actions of symmetric group $\sigma_n$ on $P(n)$ for each $n$


such that they satisfy associativity, unitality and equivariance axioms.
Here I am not sure about what this "action of a symmetric group" means.
I know the definition of action of symmetric group $\sigma_n$ on a set $X$: it is any function $\cdot: \sigma_n \times X \to X$ such that $g\cdot(h\cdot x) = (gh)\cdot x$ and $e\cdot x = x$.
However I do not know what they mean in the definition of symmetric operad. What does an action of a symmetric group on an object of some symmetric monoidal category $V$ mean?
 A: It means that for all $\sigma \in \Sigma_n$, you must give a morphism $f_\sigma : P(n) \to P(n)$, such that for all $\sigma, \tau$, $f_\sigma \circ f_\tau = f_{\tau \sigma}$ (with the usual definition of operads, this is an action on the right so it's reversed) and $f_1 = \operatorname{id}$. You can use the notation $f_\sigma(x) = x \cdot \sigma$ so that the conditions above become $(x \cdot \tau) \cdot \sigma = x \cdot (\tau \sigma)$ and $x \cdot 1 = x$, as usual, although the notation doesn't mean much in a general category.
Equivalently, this is a morphism of groups $f : \Sigma_n^{op} \to \operatorname{Aut}(P(n))$ (again, you must reverse the group structure, because again, usually it's a right action, not a left action).
Also equivalently, it's a morphism of monoids $f : \Sigma_n^{op} \to \operatorname{Hom}(P(n), P(n))$. Indeed, if you have such a morphism of monoids, then $f(\sigma)$ is automatically an automorphism of $P(n)$ for all $\sigma$, because $f(\sigma) \cdot f(\sigma^{-1}) = \operatorname{id} = f(\sigma^{-1}) \circ f(\sigma)$.

Really, this is nothing esoteric and it's certainly what you intuitively think an "action of a group" is. All these definitions are equivalent, so you can use the one that you prefer. If you're not sure why they are equivalent, try to prove it.
If you want a more arcane approach to this, you can say that any locally small category is enriched in $\mathsf{Set}$ (this is almost tautological) and go on from there. Note also that it's not really a question about operads, other than the fact that in the definition of an operad, the action is on the right (but it doesn't change much). Nor is it even special to symmetric monoidal categories.
