$\left(n^n\right)_b = \left(n\right)_b\left(n\right)_b\ldots\left(n\right)_b$ A friend of mine asked me this today and I was not able to give him an answer. Given a base $b$, find (or show the absence of) an integer $n>1$ s.t.
$$\left(n^n\right)_b=\left(n\right)_b\left(n\right)_b\ldots \left(n\right)_b.$$
The notation $\left(n\right)_b \left(m\right)_b$ means concatenating the representation of the number $n$ in base $b$ with the representation of the number $m$ in base $b$. The $\ldots$ in the above should be taken to mean "any number of" consecutive $\left(n\right)_b$.

Edit:
@alex.jordan points out the interesting relaxation (see comments):
$$\left(n^n\right)_b=\left(\underbrace{0...0}_k\,n\right)_b\left(\underbrace{0...0}_k\,n\right)_b\ldots \left(\underbrace{0...0}_k\,n\right)_b.$$
where $\underbrace{0...0}_k$ should be understood to mean $k\geq0$ leading zeros.
An answer to either is acceptable. The construction of such an $n$ giving an affirmative answer for the first question will (naturally) give an affirmative one to the second with $k=0$.

Edit 2:
The general problem is prohibitively difficult. I will accept an answer for $b=10$.
 A: If
$$
(n)_b = (d_1d_2\ldots d_t)_b \\
(n^n)_b = (n)_b(n)_b\cdots(n)_b
$$
for some digits $d_1,d_2,\ldots,d_t$ in base $b$, then $(n)_B$ is a single digit and
$$
(n^n)_B=(n)_B(n)_B\cdots(n)_B
$$
is a repeated single digit base $B=b^t$. This is also true if some $d_i=0$, allowing for leading zeros.
So for the example given by @alex.jordan in the comments $3^3=011011_2$ is equivalent to $3^3=33_8$.
Thus it suffices to look for solutions where $n<B$ is represented by a single base $B$ digit. In that case we have
$$
n^n = nB^{l-1} + nB^{l-2} + \cdots + nB + n \\
n^{n-1} = \frac{B^l-1}{B-1}
$$
For $l=1$ there are no solutions with $n>1$. For $l=2$ we get the pattern noted by @PeterKošinár in the comments, solutions with $B = n^{n-1}-1$.
For $l\ge3$ we have $(B-1)n^{n-1} + 1 = B^l$. Let $a=1,~b=(B-1)n^{n-1},~c=B^l$ then
$$
a + b = c \\
\operatorname{rad}(abc) \le nB(B-1) < l B^2 \log B < \left(B^l\right)^{1/(1+\epsilon)} \text{ for some }\epsilon>0,n>3
$$
where $\operatorname{rad}(abc)$ is the product of distinct primes dividing $abc$. Assuming the abc conjecture there can only be finitely many pairs $n,B$ like this, so there generally won't be a solution for a given $B$. In fact I'd conjecture there are no solutions like this, but am not able to rule out exceptions.
Now if there is a solution with $n^n = n(B+1)$ and $B=b^t$ with $n>2,t>1$ then it can also be a solution in base $b$ only if $n^{n-1}=b^t+1$. By Catalan's conjecture (Mihailescu's theorem) the only powers differing by $1$ are $8,9$, so conditioned on my conjecture of the previous section $3^3=011011_2$ would be the only multi-digit solution.
