A rep-digit number is of the form $$b\cdot \frac{10^a-1}{9}$$ with an integer $b$ with $1\le b\le 9$ and a positive integer $a$. In the decimal expansion , digit $b$ occurs $a$ times and no other digit.
Conjecture : A rep-digit number cannot be a Carmichael-number
I checked upto $a=5\ 000$ ( $5\ 000$ digits ) without finding a Carmichael-number of this form (in fact , not even a Poulet-number). Can we prove that there actually is none ?
We can rule out $b=9$ and $b$ even since a Carmichael number must be odd and squarefree and $b=9$ implies divisor $9$.
Disclaimer: This is not a proof, just a long comment that may result in one.
We know that $b$ cannot be even, as there are no even Carmichael numbers. Also, since Carmichael numbers are squarefree we know that $b$ can never be $9$. Thus, we have $b \in \{ 1,3,5,7\}$.
First, define $$f_b(a) = b \cdot \frac{10^a-1}{9}\,.$$
Now, let us explore $b=5$, we know that the rep-digit $f_5(a)$ is divisible by $5$, for all $a$. Hence, if it is a Carmichael number, $f_5(a)-1$ must be divisible by $4$, therefore we must have $a = 1$ (you can see this holds by using the divisibility trick upon dividing by $4$; you only consider the last two numbers). Since $a=1$ gives $f_5(1)=5$, a non-Carmichael number we know that $b$ cannot equal $5$. Hence $b \in \{ 1,3,7\}$.
If $b=7$, we know that the rep-digit $f_7(a)$ is divisible by $7$, for all $a$. Hence, if it is a Carmichael number, $f_7(a)-1$ must be divisible by $6$, therefore we must have $a \equiv 1 \pmod 3$. Also, $f_7(a)$ must be squarefree in order to be a Carmichael number, thus it can not be divisible by $49$, giving that $a \not\equiv 0 \pmod 6$. If $a \equiv 0 \pmod 2$, we know that $f_7(a)$ is divisible by $11$, thus $f_7(a)-1$ must be divisible by $10$, which is impossible. Hence $a \not\equiv 0 \pmod 2$.
If $b=3$ and $a \equiv 0 \pmod 3$, we have that $f_3(a)$ is divisible by $9$. Thus $a \not \equiv 0 \pmod 3$. As the same argument for $a\equiv 0 \pmod 2$ holds as above, we must have $a \not \equiv 0 \pmod 2$. Thus we must have $(a \bmod 6) \in \{ 1,5 \}$
Now, for $b=1$, things get more difficult. The argument for the divisibility be $11$ does not eliminate cases. However, due to the number needing to be squarefree (note that these relations also hold for $b=3$ and $b=7$), we still know that $a \neq 0 \pmod 9$ because of $3^2$, $a \neq 0 \pmod {22}$ because of $11^2$, and $a \neq 0 \pmod {42}$ because of $7^2$.
For divisibility reasons, we have $f_1(a)$ is divisible by $13$ if $a \equiv 0 \pmod 6$ and $f_1(a)-1$ cannot be divisible by $12$ (unless $a=1$, which is no Carmichael number), thus $a \not\equiv 0 \pmod 6$. Also, because of divisibility by $37$ we have that $a \not \equiv 0 \pmod 3$. Divisibility by $101$ gives $a \not \equiv 0 \pmod 4$. Divisibility by $41$ gives $a \not \equiv 0 \pmod 5$. Divisibility by $4649$ gives $a \not \equiv 0 \pmod 7$. Divisibility by $21649$ gives $a \not \equiv 0 \pmod {11}$. Divisibility by $53$ gives $a \not \equiv 0 \pmod {13}$. I could continue the list, but I think this is enough to encourage a different type of search to be made.
Most of these divisibilities also hold for the cases $b=3$ and $b=7$.
In conclusion, we must have $b \in \{1,3,7\}$.