Growth of $\frac{1}{m-1} \sum_{t | m-1} t \varphi(t)$ I am interested in bounds for 
$$S_m = \frac{1}{m-1} \sum_{t | m-1} t \varphi(t)$$
as $m$ gets large. 
 A: If you let $n = m-1$, and $f(n) = S_{m} / n$,
 then $f(n)$ is a multiplicative function, with Euler product
$$\frac1{n^2}\prod_{p^r \mathop{\mid\mid}n} \sum_{k=0}^r p^k\phi(p^k) = \prod_{p^r \mathop{\mid\mid}n} p^{-2r}\left(1+p(p-1)+p^3(p-1)+\cdots+p^{2r-1}(p-1)\right).$$
This simplifies down to $$f(n) = \prod_{p^r \mathop{\mid\mid}n} \left(\frac{p+p^{-2r}}{p+1}\right) \le \prod_{p \mid n} \left(\frac{p+1/4}{p+1}\right).$$
Since $\sum_p 1/p$ diverges, this product can be made arbitrarily close to $0$ by choosing $n$ to be divisible by many many small primes.  So no, $S_m$ cannot be bounded below by a linear function of $m$.
On the other hand, since $f(n) \ge n/\sigma(n)$ (where $\sigma$ is the sum of divisors function), it is possible to bound $S_m$ from below by $m / (2\log \log m)$ for large enough $m$.  The slow growth of $\log \log m$ explains your empirical observations.
A: Proceeding as in the answer above let $T(n) = S(n+1)$ and we'll examine $T(n)$ in more detail. Write it as follows:
$$ T(n) = \frac{1}{n} \sum_{t|n} t \varphi(t) =  
\sum_{t|n} \varphi(t) \left( \frac{n}{t} \right)^{-1}.$$
It now follows from basic properties of the Euler totient that
$$ L(s) = \sum_{n\ge 1} \frac{T(n)}{n^s} = \frac{\zeta(s-1)}{\zeta(s)}\zeta(s+1).$$
For certain Dirichlet series of which this is an instance, we have for $c$ chosen correctly that $$ T(n) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} L(s) n^{s-1} ds.$$
Now taking the residues at $s=2$ and $s=0$, we obtain
$$ T(n) \sim 1/6\,{n}^{-1}+6\,{\frac {\zeta  \left( 3 \right) n}{{\pi }^{2}}} $$
which confirms your hypothesis that on average $n$, $T(n)$ is linear. This expansion could be continued and that is where the necessary fluctuation will appear.
We can also apply the Wiener-Ikehara theorem to study the average order of the sum of $T(n)$, which is much more smooth. We have
$$\frac{1}{n} \sum_{k=1}^n T(n) = \frac{1}{2n} T(n)+
\frac{1}{n}\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} L(s) n^s/s ds,$$
which gives
$$\frac{1}{n} \sum_{k=1}^n T(n) \sim \frac{1}{2n} T(n)+\frac{1}{n}
\left( 1/6\,\ln  \left( n \right) +1/6\,\gamma-2\,\zeta  \left( 1,-1 \right) -1/6\,\ln
 \left( 2\,\pi  \right) +3\,{\frac {\zeta  \left( 3 \right) {n}^{2}}{{\pi }^{2}}}\right).$$
This last approximation is of truly stunning accuracy, which is why I include it here even if doesn't directly answer the OP's question.
A: We can certainly bound it from above since $$\frac{1}{m-1}\sum_{t|m-1}t\varphi(t) \leq \frac{m-1}{m-1}\sum_{t|m-1}\varphi(t) = m-1$$
