Center of the category of special $\lambda$-rings 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).
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 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 $\mathrm{Z}(\lambda\mathrm{Ring})$?
 A: 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.
