Suppose I have X ∼ Bin(1,p). Is the distribution of X^2 the same as X?
I know that X+X ∼ Bin(1+1,p), which would lead me to the (probably incorrect) assertion that X^2 ∼ Bin(1*1,p).
I also know that if for Bin(n,p), where n = 1, then only one trial happens. Is this the only situation in which X^2 has the same distribution as X?
Yes, when $X$ is a Bernoulli random variable $X^2$ has the same distribution; it has the same support $\{0,1\}$ and realises each outcome with the same probability. Moreover, as Robert comments, it is not just the same distribution, this makes it the same random variable.
However, $X^2$ does not generally have the same distribution as $X$ when $X$ is the count of successes in a series of iid Bernoulli trials. This can be seen by examining the supports, as $X$ has the support of $\{0,1,2, ..., n\}$ while $X^2$ is supported on $\{0,1,4, ..., n^2\}$. Only when $n=1$ are these identical.
Further $2X\nsim\mathcal{Bin}(2,p)$ for the same reason.
If you had an independent and identically distributed variable, $Y\sim\mathcal {Bin}(1,p)$, then $X+Y\sim\mathcal{Bin}(2,p)$. But $X$ is clearly not independent from itself.