How is an exchange rate process a martingale under any measure?
Suppose a process for a stock price of a US-based company traded in the USA is, under the USD money-market numeraire:
$$ dS_t=S_tr_{USD}dt+S_t\sigma_SdW_1(t) $$
Using fundamental theorem of asset pricing, we must have that:
$$ \frac{S_0}{B_{USD}(t_0)}=\frac{S_0}{1}\stackrel{?}{=}\mathbb{E}^Q_{USD}\left[\frac{S_t}{B_{USD}(t)}\right]=\mathbb{E}^Q_{USD}\left[\frac{S_0e^{r_{USD}t-0.5\sigma_S^2t+\sigma_SW_1(t)}}{e^{r_{USD}t}}\right]=S_0 $$
So clearly, the discounted process for $ S_t $ is a martingale under $ B_{USD}(t) $ as numeraire.
Suppose I am interested in the exchange rate between USD and EUR, and I denote the process that describes how many units of USD I need to pay for 1 unit of EUR as $ X_t $ (i.e. analogously to the process $ S_t $ , which tells me how many units of USD I need to pay for 1 unit of $ S_t $ ).
Let the process of $ X_t $ be as follows:
$$ dX_{EUR\rightarrow USD}(t)=(r_{USD}-r_{EUR})X_{EUR\rightarrow USD}(t)dt+\sigma_XX_{EUR\rightarrow USD}(t)dW_2(t) $$
The Forward on $ X_t $ is denoted as $ F(X_t)=\mathbb{E}_{USD}^Q[X_t|X_0] $ .
The no-arbitrage condition on the forward is trivially: $$ F(X_{EUR\rightarrow USD}(t))=\frac{e^{r_{USD}t}}{e^{r_{EUR}t}}X_{EUR\rightarrow USD}(t_0) $$
Clearly, this condition is satisfied because $$ \mathbb{E}{USD}^Q[X_t|X_0]=\mathbb{E}{USD}^Q[X_0e^{r_{USD}t-r_{EUR}t-0.5\sigma_X^2t+\sigma_XW_2(t)}]=X_0e^{r_{USD}t-r_{EUR}t}=\frac{e^{r_{USD}t}}{e^{r_{EUR}t}}X_{EUR\rightarrow USD}(t_0) $$
But clearly, under the USD numeraire, the discounted process for $ X_t $ is NOT a martingale, since:
$$ \frac{X_0}{1}\stackrel{?}{=}\mathbb{E}^Q_{USD}\left[\frac{X_t}{B_{USD}(t)}\right]=\mathbb{E}^Q_{USD}\left[\frac{X_0e^{r_{USD}t-r_{EUR}t-0.5\sigma_X^2t+\sigma_XW_2(t)}}{e^{r_{USD}t}}\right]=X_0e^{-r_{EUR}t}\neq X_0 $$
So the process for $ X_t $ cannot be a valid process under the $ B_{USD}(t) $ numeraire. It would not be a valid process under the $ B_{EUR}(t) $ numeraire either, because again, if discounted by the EUR numeraire, it would not be a martingale.
Where is the catch here?
The catch is that when a stock pays dividends, say, by a continuously compounded dividend yield $ q $ then $$ \frac{dS_t}{S_t}=r_{USD},dt\color{red}{-q,dt}+\sigma,dW_t,, $$ and $ S_te^{-r_{USD}t} $ is not longer a martingale either. Thank god this can be fixed because when you add to this the present value of the paid dividends then the process $$ M_t=S_te^{-r_{USD}t}+\int_0^tqS_ue^{r_{USD}(t-u)},du $$ is a martingale. See this answer.
Now to the FX rate. The well-known SDE $$ \frac{dX_t}{X_t}=r_{USD},dt\color{red}{-r_{EUR},dt}+\sigma,dW_t $$ suggests that $ r_{EUR} $ is the continuous compounded dividend yield that you receive when you hold one unit of $ X_t $ which is the price of one EUR in USD.
This makes a lot of sense because when we have 1 EUR in cash we can say that we hold $ X_t $ USD and we do receive $ r_{EUR} $ per time and share in dividends from holding that cash.
To make a long story short: $ X_te^{-r_{USD}t} $ is not a martingale but $$ M_t=X_te^{-r_{USD}t}+\int_0^tr_{EUR}X_ue^{r_{USD}(t-u)},du $$ is one.