For which RSA moduli, precisely, is $e=d$ for all $e$? This question shows that there are at least two valid RSA moduli $n$, namely $35$ and $91$, such that for any $e$ coprime to $\lambda(n)$, $$e^2\equiv1\mod\lambda(n)\text.$$
Reading the linked question, I wondered if there is a "simple" description of the set of numbers $n$ that have this property. In particular, are there infinitely many such numbers?
I wondered if there is a "simple" description of the set of numbers n that have this property.


Yes, there is; $n$ has a prime factorization $p_1 \cdot p_2 \cdot ... \cdot p_n$ such that all the primes are unique (i.e. $n$ is square-free), and for each prime factor $p_i$, $p_i-1$ must be a divisor of 24.  In other words, each prime must be a member of the set $\{2, 3, 5, 7, 13\}$

Why is this?  Well, $e^2 \equiv 1 \mod \lambda(n)$ (for $e$ r.p. to $\lambda(n)$) is equivalent to $e^2 \equiv 1 \mod p_i-1 $ for all $i$ (for $e$ r.p. to $p_i-1$); this latter holds only if $p_i-1$ is a divisor of 24.

As a consequence, we see that there is only a finite number of such modulii; if we restrict ourselves to modulii with two factors, we see that 91 is the largest possible.  If we include multiprimes, then the largest possible is $2730 = 2 \times 3 \times 5 \times 7 \times 13$