If $A$ is similar to an upper triangular matrix, must A also be upper triangular? Say $A$ is an $n \times n$ matrix, I understand stand that 2 matrices $A, B$ are similar iff there exist an invertible matrix $Q$ such that $A =$ $Q^{-1}BQ$,
so if $B$ is already an upper triangular matrix, does $Q^{-1}BQ$ affects B and make it not upper-triangular ?
 A: Over any field, if $A$ is a lower triangular matrix, then $B=PAP^{-1}$ is upper triangular when $P=P^{-1}$ is the reversal matrix obtained by flipping the identity from left to right. So, the answer to your question in the title is no.
As for your other question, if you pick an arbitrary $Q$, it may of course break triangularity. Yet, in triangularisation, the $Q$ we pick is not arbitrary but depends on the matrix to be triangularised. E.g. when $B$ is already triangular, you can simply choose $Q=I$. Then $A(=B)=Q^{-1}BQ$ is triangular.
A: Over the field $\mathbb{C}$, by Schur's decomposition $A$ need not be upper triangular for it to be similar to an upper triangular matrix.
To generalize to other fields, as stated in the comments: if a matrix is diagonalizable then it is similar to an upper triangular matrix. So again no, $A$ need not be upper triangular.
Here are some other interesting posts/articles: Conditions for Schur decomposition and its generalization
http://www.sciencedirect.com/science/article/pii/S0024379597800079 
