Recall that the center $\mathrm{Z}(C)$ of a category $C$ is the monoid of endomorphisms of $\mathrm{id}\_C$. Thus $\eta \in \mathrm{Z}(C)$ is given by a familiy of endomorphisms $\eta\_x : x \to x$, where $x \in C$, such that for all morphisms $x \to y$ the obvious diagram commutes. The center of the category of rings is trivial (see [here](https://mathoverflow.net/questions/61197/center-of-the-category-of-r-algebras)).


In the category of special $\lambda$-rings $\lambda\mathrm{Ring}$, we have for every $q \in \mathbb{N}\_{\geq 1}$ the Adams-Operation $\Psi^q$ which is known to be a $\lambda$-ring endomorphism for every special $\lambda$-ring and is compatible with $\lambda$-ring homomorphisms (see [these notes](http://sites.google.com/site/darijgrinberg/lambda) by Darij Grinberg). Besides, we have $\Psi^p \circ \Psi^q = \Psi^{pq}$ and $\Psi^1 = \mathrm{id}$. This shows that there is a homomorphism of monoids


$(\mathbb{N}\_{\geq 1}, \*) \to \mathrm{Z}(\lambda\mathrm{Ring})$, $q \mapsto \Psi^q$


It is easy to see that it is injective. It is also surjective? If not, what do we have to add to get $\mathrm{Z}(\lambda\mathrm{Ring})$?



Center of the category of special $\lambda$-rings
Your map is surjective too.


The free (special) $\lambda$-ring on one generator is a polynomial algebra of the form $F=\mathbb{Z}[\lambda^1(x),\lambda^2(x),\lambda^3(x),\dots]$. (This is well-known; I think Donald Yau proves it in his book on $\lambda$-rings.) The set of endomorphisms of the forgetful functor $\lambda\mathrm{Ring}\to \mathrm{Set}$ corresponds to the underlying set of $F$, so $Z(\lambda\mathrm{Ring})\subset F$.


One way to proceed from here is to use the fact that $A\mapsto A\otimes \mathbb{Q}$ is a functor $\lambda\mathrm{Ring}\to \mathbb{Q}\backslash\lambda\mathrm{Ring}$, and that $\lambda$-rings containing $\mathbb{Q}$ are nothing more that commutative $\mathbb{Q}$-algebras equipped with Adams operations. It should be easy to see that 
$Z(\lambda\mathrm{Ring})\subseteq Z(\mathbb{Q}\backslash \lambda\mathrm{Ring})$, since $F\subseteq F\otimes\mathbb{Q}$, and that $Z(\mathbb{Q}\backslash\lambda\mathrm{Ring})=\mathbb{N}$.


**Added.**


Let me write $F\{x\_1,x\_2,\dots\}$ for the free (special) $\lambda$-ring on generators $x\_1,x\_2,\dots$. Then $F\{x\_1,x\_2\}\approx F\otimes F$, since coproducts in $\lambda$-rings are tensor products. There is a comultiplication $\Delta\colon F\{x\}\to F\{x\_1,x\_2\}$ (i.e., $F\to F\otimes F$) defined by sending $x\mapsto x\_1+x\_2$; it makes $F$ into a Hopf algebra. The map $\Delta$ encodes how polynomials in $\lambda$-operations act on sums, so we see that $\psi^k(x)\in F$ is primitive: $\Delta(\psi^k(x))=\psi^k(x\_1)+\psi^k(x\_2)$. 


The subgroup $P\subset F$ corresponds precisely to the set of polynomials $f(\lambda^1,\lambda^2,\dots)$ such that $f(x+y)=f(x)+f(y)$ in any $\lambda$-ring.
I want to identify $P$ with the $\mathbb{Z}$-linear span of the $\psi^k(x)$. It is a little easier to identify the subgroup of primitives in $F\otimes \mathbb{Q} \approx \mathbb{Q}[\psi^1(x),\psi^2(x),\psi^3(x),\dots]$ as the $\mathbb{Q}$-linear span of the $\psi^k(x)$ (for instance, using structure theorems for Hopf algebras; $F\otimes\mathbb{Q}$ is primitively generated as a Hopf algebra.). 


There is a second comultiplication $\Delta'\colon F\{x\}\to F\{x\_1,x\_2\}$ sending $x\mapsto x\_1x\_2$, which encodes how operations act on products. We want the elements inside $P$ (or just $P\otimes \mathbb{Q}$) which are grouplike with respect to $\Delta'$, (i.e., $\Delta'(u)=u\otimes u$). We already know that the $\psi^k(x)\in P$ have this property, so we just need to show that if a linear combination of $\psi^k(x)$'s is grouplike wrt to $\Delta'$, then it is just a single $\psi^k(x)$, which is relatively elementary.


