For which $t \in \mathbb{N}$ does $\varphi(t) \mid t$? Just curious about this one. Don't really know where to start, any help would be appreciated thanks.

Clearly, it's true for $t = 1$. Consider $t > 1$.
Let $t = p_1^{k_1}\cdots p_n^{k_n}$ be the prime factorisation of $t$. Then, we have
$$\varphi(t) = t\left(1 - \dfrac{1}{p_1}\right)\cdots\left(1 - \dfrac{1}{p_n}\right).$$
Thus, you would require
$$\left[\left(1 - \dfrac{1}{p_1}\right)\cdots\left(1 - \dfrac{1}{p_n}\right)\right]^{-1}$$
to be an integer. The above can be written as
$$\dfrac{p_1}{p_1 - 1}\cdots\dfrac{p_n}{p_n - 1}.$$
If $n = 1$, then we just have $p_1/(p_1 - 1)$. This clearly forces $p_1 = 2$. (Else, the denominator would be even and numerator odd.) Thus, all numbers of the form $2^a$ for $a \ge 1$ work.  
Now, if $n = 2$, then we have $\dfrac{p_1}{p_1 - 1}\dfrac{p_2}{p_2 - 1}$.
Once again, one of the primes is forced to be $2$. Say $p_1 = 2$. Then, the expression simplifies to
$$\dfrac{2p_2}{p_2 - 1}.$$
Note that $p_2 - 1$ and $p_2$ are coprime and thus, we must have $p_2 - 1 \mid 2$. This forces $p_2 = 3$. Thus, we have numbers of the form $2^a3^b$ where $a, b \ge 1$.
If $n > 2$, then we have no solutions. The denominators would have at least two powers of $2$ whereas the numerator could have at most 1.
To conclude we have the following set:
$$t \in \{1\} \cup \{2^a : a \ge 1\} \cup \{2^a3^b : a, b \ge 1\}.$$
(This could probably be written more concisely but this gives better clarity. Note that numbers of the form $3^b$ for $b \ge 1$ are not solutions.)
