Normality result using Schur form 
Let $A\in\mathbb{C}^{n\times n}$ have eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_n$ Using Schur Form so that
\begin{equation}\sum_{i,j=1}^{n}|a_{ij}|^2=\sum_{i=1}^{n}|\lambda_i|^2\implies A\text{ normal}\end{equation}
Using this result; Prove that 2 unitarily similar matrices $A=(a_{ij})$ and $B=(b_{ij})$ satisfy the condition
\begin{equation}\sum_{i,j=1}^n|a_{ij}|^2=\sum_{i,j=1}^n|b_{ij}|^2\end{equation}
and therefore show the converse of the first statement is true, that is $A$ normal $\implies \sum_{i,j=1}^n|a_{ij}|^2=\sum_{i}^n|\lambda_i|^2.$

I have carried out the following working however I am not sure I am using the correct approach to this problem and have somewhat hit a wall on what to do.

We know by Schur form that we can have $A=UTU^*$ for some unitary $U$ and a triangular $T$ with $t_{ii}=\lambda_i.$ Notice that $trace(A^*A)=\sum_{i,j=1}^n|a_{ij}|^2$, and $A^*A=UT^*TU^*,$ at this point i have $trace(UT^*TU^*)=trace(U^*UT^*T)=trace(T^*T)=\sum_{i,j=1}^n|t_{ij}|^2,$ but i cant see how to get this to be the eigenvalues given it isnt the diagonal entries in the summation, or how to get this implying that $A$ is normal.

I think the above working might be able to be applied to the proof regarding the condition for unitary similar matrices, but I am unsure on how to tackle the rest of the problem.

You are on somewhat of a right track. The point is that for a triangular matrix the eigenvalues are on the diagonal, so that the sum of squares of the eigenvalues is equal to the sum of squares of all entries if and only if there are no other entries, i.e. if the matrix is diagonal. So now we can apply this:
If $\sum_{i} \lambda_i^2=\sum_{i,j} |a_{ij}|^2$  then we get
$$\sum_i t_{ii}^2=\sum_{i} \lambda_i^2=\sum_{i,j} |a_{ij}|^2= \sum_{i,j} |t_{ij}|^2$$
so $T$ is diagonal and $A=UDU^*$, is normal.
Conversely, if $A$ is normal then (by diagonalizations theorem for normal matrices) $A=UDU^*$ and
$$\sum_{i,j} |a_{ij}|^2=\sum_{i,j} |d_{ij}|^2=\sum_{i} d_{i,i}^2=\sum_{i} \lambda_i^2$$
