Note that $\DeclareMathOperator\End{End}\End(C)$ is the category of functors $\newcommand{\BN}{{\mathrm B\mathbb N}}\renewcommand\hom{\operatorname{hom}}\hom(\BN,C)$, where $\BN$ is the category with one object and $\mathbb N$ morphisms, also called the *walking endomorphism*. Thus $\End(\End(C)) = \hom(\BN^2,C)$, by the hom-tensor adjunction for categories.

Iterating, let $\BN^{\oplus\infty}$ denote the category with one object, whose morphisms are the abelian monoid (under addition) of finite (but arbitrarily long) sequences of nonnegative integers. There is an isomorphism $\BN \times \BN^{\oplus\infty}$, giving an isomorphism of categories $\End(\hom(\BN^{\oplus\infty},C)) = \hom(\BN^{\oplus\infty},C)$. So this gives a class of examples.

There are other examples. For instance, I could have used $\BN^{\times\infty}$, the countable direct product of $\BN$ with itself, instead of $\BN^{\oplus\infty}$. Then I'd still have $\BN \times \BN^{\times \infty} = \BN^{\times \infty}$, and so $\End(\hom(\BN^{\times\infty},C)) = \hom(\BN^{\times\infty},C)$ for any category $C$.

I had hoped to prove that the first class of fixed points was universal. Here is the argument that I had hoped to use: Suppose we are given an equivalence of categories $F: \hom(\BN,D) \to D$. Then by currying we do get a sequence of equivalences $F_k: \hom(\BN^{k+1},D) \to \hom(\BN^k,D)$ for any finite $k$. We can try to compute the limit of this system in two different ways. On the one hand, it is a system of equivalences, and so its limit is equivalent to every other term in the system. On the other hand, $\BN^{\oplus\infty}$ is a colimit, and colimits in the first variable of $\hom(,)$ become limits, and so it is very tempting to think that $\underset{\leftarrow k}{\lim\limits} \hom(\BN^k,D) = \hom(\BN^{\oplus \infty},D)$. In this way, you could hope to conclude that $D \simeq \hom(\BN^{\oplus\infty},D)$.

Unfortunately, I think that this argument has no chance of working. (The following is based on a conversation with William Johnson.) For, suppose that it did. Then there would be, for any category $C$, a natural equivalence of categories $\hom(\BN^{\oplus \infty}\times \BN^{\times \infty},C) \simeq \hom(\BN^{\times \infty},C)$. By the Yoneda lemma, this would imply that $\BN^{\oplus\infty} \times \BN^{\times \infty} \simeq \BN^{\times \infty}$, or, equivalently, $\newcommand\NN{\mathbb N}\NN^{\oplus\infty} \times \NN^{\times \infty} \cong \NN^{\times \infty}$ as commutative monoids.

I claim that this is impossible. It is easier to work with groups, so let's allow subtractions. Then I claim that $\newcommand\ZZ{\mathbb Z} \ZZ^{\oplus \infty} \times \ZZ^{\times \infty} \not\cong \ZZ^{\times \infty}$. The proof is as follows:

Suppose that you have any linear map $\ZZ^{\oplus \infty} \to \ZZ^{\otimes \infty}$. You can write this as an $\infty\times \infty$ matrix, where the first column is where $e_1 = (1,0,\dots)$ goes, the second column is where $e_2 = (0,1,0,\dots)$ goes, and so on. (The matrix may have non-zero entries everywhere, if you want.) I will try to implement the usual Gaussian elimination algorithm to diagonalize the matrix.
Suppose temporarily that the upper left corner of the matrix is $1$. Then by modifying $e_i \mapsto e_i' = e_i - \# e_1$, where $\#$ is some matrix entry, I can clear the rest of the first row. This manipulation is simply the precomposition of the matrix with some automorphism of $\ZZ^{\oplus\infty}$.

Similarly, arbitrary permutations of the columns are also implemented by automorphisms of $\ZZ^{\oplus\infty}$. What happens if the first row never has a $1$ in it? Well, $\ZZ$ is a PID; let $g$ denote the principal generator of the ideal ideal generated by all the entries in the first row, i.e. the GCD of the first row. But $g$ is by definition some finite linear combination of the entries in the first row, so again basic column operations allow us to get $g$ into the upper left corner. Finally, every entry in the first row is divisible by $g$, and so we can use elementary column operations to clear the rest of the first row.

Now look at the second row. Don't worry about the first column, but clear the rest. Etc. At the end of the day, you get some matrix $X : \ZZ^{\oplus\infty} \to \ZZ^{\times \infty}$ in column-echelon form.

Let $f_n = (1,2,6,\dots,n!,0,0,\dots) \in \ZZ^{\oplus \infty}$. This gives a sequence of terms $Xf_1,Xf_2,\dots \in \ZZ^{\times \infty}$. Because $X$ is in column-echelon form, it is clear that this sequence has a limit $Xf_\infty \in \ZZ^{\times \infty}$ (indeed, the first $n$ terms of $Xf_\infty$ agree with the first $n$ terms of $Xf_n$).

Now look at the image $[Xf_\infty]$ of $Xf_\infty$ in the quotient $\ZZ^{\times \infty} / (X \ZZ^{\oplus\infty})$. I claim first that $[Xf_\infty] \neq 0$. Indeed, suppose that $Xf_\infty = Xh \in \ZZ^{\oplus \infty}$ for some finite sequence $h\in \ZZ^{\oplus \infty}$. Say that $h$ is all zeros after the first $m$ terms. Recall that a *pivot* in $X$ is the first non-zero entry in any column. Let $n$ by the row for the $m$th pivot. Since $X$ is an injection, we can reconstruct $h$ from the first $n$ entries of $Xh$. But these first $n$ entries are precisely the first $n$ entries of $f_n$, and so $h$ and $f_n$ agree to their first $m$ spots, i.e. $h = f_m$. But again since $X$ is an injection, $X f_m \neq X f_{m+1}$, differing in the $(n+1)$th spot. So this proves that $Xf_\infty \not\in \ZZ^{\oplus\infty}$.

On the other hand, I claim that $[Xf_\infty]$ is infinitely divisible in the abelian group $\ZZ^{\times \infty} / (X \ZZ^{\oplus\infty})$. Indeed, it is divisible by $n$, because $f_{N} - f_n$ is for $N > n$, and $[Xf_\infty] = [X(f_\infty - f_n)]$. We have therefore constructed a non-zero infinitely-divisible element of $\ZZ^{\times \infty} / (X \ZZ^{\oplus\infty})$.

On the other hand, $(\ZZ^{\oplus \infty} \times \ZZ^{\times \infty})/ \ZZ^{\oplus\infty} = \ZZ^{\times \infty}$ does not contain any non-zero infinitely-divisible elements. Therefore $\ZZ^{\oplus \infty} \times \ZZ^{\times \infty} \not\cong \ZZ^{\times \infty}$, completing the proof.