Let $r=\gcd(k,e+1)$, and $p$ a prime. Then $\sigma_k(p^e) \equiv r\frac{p^{e+1}-1}{p^r -1} \bmod \sigma(p^e)$. Also, $r=1$ if and only if $\sigma(p^e)$ divides $\sigma_k(p^e)$. Thus for $k$ coprime to $\tau(n)$, we have $\sigma(n)$ divides $\sigma_k(n)$. The relation also suggests that for a given $n$ the sequence $\sigma_k(n)\bmod \sigma(n)$ is periodic in $k$ with a period dividing $L$, the least common multiple of ($1+$ each exponent) in the prime factorization of $n$. **Edit 2016.01.04**: Once can show a nonreduced representation $\sigma_k(n) = a_k\sigma(n)/b_k$ where the $b_k$ are integers not necessarily coprime to the integers $a_k$ or to $a_k\sigma(n)$, with the property that the $b_k$ are bounded and periodic with period $L$. This is not enough to show $\sigma_k(n) \bmod \sigma(n)$ is periodic with small period, unfortunately. **End Edit 2016.01.04**.

If now $n$ is multiperfect (so $n$ divides $\sigma(n)$) we have $n$ divides $\sigma_k(n)$ for $k$ coprime to $\tau(n)$. In particular if $\tau(n)$ is a power of $2$, then $n$ divides $\sigma_k(n)$ for all odd $k \gt 0$.

It is still possible that $n$ can divide $\sigma_k(n)$ for $k$ not coprime
to $\tau(n)$. However if $L$ is not prime, it seems likely that there will be more than one nonzero value of $\sigma_k(n) \bmod \sigma(n)$. If this is so, it would be one ingredient in a proof that 6 is the unique number having the titled properties, the other ingredient being that 6 is the only nontrivial multiperfect number with $L$ a prime.

**Edit 2016.01.10:** I botched an earlier edit which claimed that 6 is the only known multiperfect number $n$ which satisfies $\sigma_2(n) \bmod n = 2$. It is true, but the analysis had some flaws. However, one expects multiperfect numbers other than 1 and 6 to be a multiple of 4; when $n$ satisfies $\sigma(n) \bmod n = 0$ and $\sigma_2(n) \bmod n = 2$, and in addition $ n \bmod 4 = 0$, then all odd prime factors of $n$ except one must occur to an even multiplicity, and the remaining odd prime factor must occur to a multiplicity of 1 mod 4 and must be a prime that is 3 mod 4. While simple, these observations say a lot about $n$ and suggest that any numbers satisfying the title congruences are rare indeed, perhaps more so than odd multiperfect numbers. **End Edit 2016.01.10**

Gerhard "Mea Culpa, Mea Maxima Culpa" Paseman, 2016.01.03