What are the odds of this seemingly astronomically impossible thing happening?? What are the odds of seeing 2 different licenses plates with identical numbers and letters but the numbers being slightly flipped ( letters remain the same). Some more background info, my girlfriend and I saw a license plate we thought was funny so we snapped a picture, and 5 days later we saw the same exact plate but the numbers were slightly switched ( went from 691 to 169) we live in a fairly big city, population over 400k, and we are very shocked and confused how this can possibly happen. I’m choosing not to disclose the first 3 letters cause it’s the internet lol. Any help at the odds of this happening are much appreciated, she’s convinced it’s god and I’m just trying to be as rational as possible lmao. 
 A: OK, first the math:
It's hard to put precise numbers on this. Even if we assume that you only are looking at license plates from the same country, license plates can come in different formats of letters and digits. So, I think it's silly to try and put precise numbers on this, but we can make some rough estimates that at least get us in the neighborhood ... or at least give us some idea of the order of magnitude. 
So, let's assume you live in country with about $100$ million license plates (for example, if all license plates have $3$ letters, followed by $3$ or $4$ digits, you get in that neighborhood of number of license places).
Now, it sounds like after you saw the first license plate, you saw a near copy: the same three letters in the same order, followed by three (different) digits, but the difference was that the digits were reordered. Since there are $6$ ways to reorder the three digits, we get $5$ near copies of the first one. Now, if we assume that every license plate is randomly distributed across the country, then for every subsequent license plate that you see, you have a $5$ in $100$ million chance of seeing a license plate like the first.
Immediately, I should point out that this assumption of random distribution is implausible. For example, I would assume that in many countries, license plates are being produced 'in order'. That is, once a license plate ABC123 is produced, the next one is probably going to be ABC124. And, it is quite possible that those subsequent license plates are produced, in batches, at physical sites that will correlate with where the cars are in the country. Thus, I would guess that there is a correlation between the license plate and where you are in the country at any time. Indeed, in the US for example, different states follow different formats for letters and digits: some states have 3 letters followed by 3 digits, which is what I basically assumed, but other have 3 digits followed by 3 letters, and some have just digits. In Wyoming, the letters of a license plate indicates the county the car is registered in, so you get a really high correlation of similar license plates there.
Again, all of this makes it really hard to put numbers on this, and it makes it real silly to put any exact numbers on this.
Still, let's go with the rough estimate that there are $5$ in $100 million cars with license plates that are near copies of the first one you saw.
Now, you say you live in a sizable city, meaning that you probably see a good number of cars on the road. Let's say that in the $5$ days past you noticed another $1000$ license plate. Again, that's just a very rough estimate. Clearly you see far more cars than that you pay attention to their license plates. Indeed, from that perspective, seeing $1000$ license plates in $5$ days sems like a lot ... but please also note that there is probably a lot of unconscious visual processing going on ... and if there is a car with a near copy, it'll more likely reach your conscious perception. So, I don't think $1000$ is too bad of an estimate.
Then, the chances of encountering one just like the funny original one equals $1$ minus the probability of not seeing any plate like it, which is $(\frac{99,999,995}{100,000,000})^{1000}$ ... which we can calculate ... but again since we're doing things roughly anyway, let's just say that you can roughly multiply the chances of $1$ license plate being a near copy ($5$ in $100$ million) by a thousand ... which works out to be a $1$ in $20,000$ chance of seeing that other license plate.
Is that an astronomical number? No. it's about the chance of rolling five $6$'s in a row with a die. Evidence of God? Certainly not. 
OK, and now for the psychology behind this. Psychologically, we treat this one event of seeing two near-identical funny license plates as one (highly improbable) event. Indeed, you may have noted that the above calculations don't really calculate the event of seeing those two specific near-copies, but rather the chances of seeing the second license plate having already seen the first one, and so the calculation seems to be off in that respect.  However, let us note that there are probably a lot more funny license plates (SEX 169: very funny, sure... but so is PLZ 123 or LOL 568 or ...). The same game can be played with digits: 123 is 'interesting', but so is 234, or 124, or 321, or .... and believe me, mathematicians can even tell you interesting things about 781. So, the probability of seeing two similar funny license plates is far, far greater than seeing the two specific ones you saw. That's why I felt justified to just calculate the chances of seeing a repeat.
And even there: suppose the one you saw was SEX 169, and later you saw SEX 619. Now, you say: "Wow, only the digits were reversed: what are the chances of that?!". However, what if the second one would have been SEX 269? Now you would have said "Wow, only one digit off! : what are the chances of that?!". The point is, the very notion of 'near copy' is vague enough that many pairs of license plates would have made you believe that something highly probably was going on (because you saw and phrased it in certain specific terms), whereas in fact the chances of something like that would have been much higher.  
Furthermore, as pointed out in the comments, we only notice weird things when they do happen, but not when they don't. For example, pick up a die and roll it five times: the probability of getting exactly the very outcomes you get and in that very order is again the chance of what happened to you with the license plates. Pretty incredible! ... except that it doesn't feel incredible at all. And this is because, most likely you'll get something 'boring' like $3,5,4,2,2$, and so this is seen as a non-event.
OK, but now think of all the opportunities for something weird to happen to you or your girlfriend ... but when nothing happens I doubt you ever say "Huh! Nothing weird happened!". Indeed, there have been many spans of five days (just three in the past week alone!) during which you could have noticed a funny license plate and a near copy ... but this never happened. Psychologically these 'non-events' never seem to enter our judgments though. It's like rolling $5$ dice many, many times .. but pretending that when you finally roll five $6$'s, that this was somehow the one and only time you rolled those dice: "Wow, what were the chances of that! It must be $\frac{1}{6}^5$!" But of course the chances of that happening at least once between all of those rolls were far higher. And again, if you would have rolled five $5$'s you also would have said "Wow, what are the chances of that?!".
So yes, I think the title of your Post is exactly right: this event was seemingly astronomically impossible ... because our psychology is just not framing this appropriately.
Finally, I see these kinds of "Wow, that's incredible!  The chances of that must be astronomically high!  Certainly God was involved!" kinds of stories all the time. If it isn't license plates, then it's person A saving person B's life, and later it turns out that some years earlier, person B had saved person's A's life, etc. etc.  But with a good number of interesting license plates , there should be plenty of opportunities for someone to see an interesting license plate and its near copy in the span of five days. In fact, with so many people, and with so many possible spans of five days, this should be a near certainty, and that's just for funny license plates! With so much wiggle room for two events to be 'similar', and with quadrillions of 'events' to take place in the world at any time, I would only start considering it evidence of God if weird coincidences would not ever happen to someone.
As Aristotle said: "Unlikely things are likely to happen!" 
A: For simplicity, let's assume that all license plates contain precisely six characters, that each character is either a letter (A-Z) or a digit (0-9), and that all permutations of these characters are allowed1.
This gives us $N_\mathrm{plates}=36^6 = 2,176,782,336$ possible license plates. Save that number for later.

Let me rephrase your question into a different, but perhaps more enlightening one2:

How many cars would you have to see in order for the probability of seeing two cars with similar license plates to be greater than 50%? 

For the purpose of this answer, let's define two license plates to be "similar" if they contain the same six characters.
Let define $A_n$ to be the probability that two similar plate exist in a collection of $n$ random plates. We want to find the minimum $n$ such that $P(A_n) > 0.5$.
To make this computation, we look at $P(A'_n)$, the probability that in a pool of $n$ plates, no two plates are similar. Since the events $A_n$ and $A'_n$ are mutually exclusive, $P(A)=1-P(A'_n)$.
If we have only one license plate, clearly no two plates match, so $P(A'_1)=1$.
For two license plates ($A'_2$), we start by observing that (ignoring repeated characters for simplicity) there are $6!=720$ possible ways to permute six characters. Recall that there are $N_\mathrm{plates}$ possible license plates in existence and therefore, for any given license plate, there are $N_\mathrm{plates}-6!$ license plates that are not similar to it.
So, the probability that the second plate is not "similar" to the first is
$$
P(A'_2) = \frac{N_\mathrm{plates}-6!}{N_\mathrm{plates}} \approx 0.9999996692
$$
Let's continue on to $n=3$. The third plate can't be similar to either the first or the second, and there are $N_\mathrm{plates} - 2\cdot6!$ possible license plate that meet this criteria. The second plate also still number not be similar to the first. Hence,
$$
P(A'_3) = \left(\frac{N_\mathrm{plates}-6!}{N_\mathrm{plates}} \right)\left(\frac{N_\mathrm{plates}-2\cdot 6!}{N_\mathrm{plates}}\right) \approx 0.9999990077
$$
You can probably already see the pattern here. In general, we can deduce that the probability that any collection of $n$ license plates has no two similar plates is given by
$$
P(A'_n) = \left(\frac{N_\mathrm{plates}-1\cdot 6!}{N_\mathrm{plates}}\right)\left(\frac{N_\mathrm{plates}-2\cdot 6!}{N_\mathrm{plates}}\right)\cdots\left(\frac{N_\mathrm{plates}-n\cdot 6!}{N_\mathrm{plates}}\right)\\
= \frac{1}{N_\mathrm{plates}} \prod_{i=1}^n \left( N_\mathrm{plates}-i\cdot 6! \right)
$$
By brute force computation, we can use this formula to find the minimum $n$ such that $P(A'_n) < 0.5$, which is $n=2048$:
$$
P(A'_{n=2048}) \approx 0.499834086459520
$$
And so,
$$
P(A_{n=2048}) \approx 0.500165913540480
$$
Therefore, in order to have a greater than 50% chance of seeing two similar plate, you'd need to look at about 2,048 cars. This number gets drastically smaller once you consider the fact that many jurisdictions put additional restriction on what license plate are allowed. For instance, if a license plates are required to have precisely three letters and three digits, then the number of plates you would need to see before finding two similar plates with 50% odds reduces to only 823!3

1 In reality, some states choose to reserve or ban some license plate combinations, which adds some more complexity to the calculation if you really cared for an absolute answer.
2 In probability, this kind of problem is known as the birthday problem.
3 Under this scheme, we have $N=26^3\cdot10^3=17,576,000$ possible plates, with each given plate being similar to $(3!)^2=36$ other plates. We then compute the minimum $n$ such that $$\frac{1}{N}\prod_{i=1}^{n} \left( N - i\cdot36\right) >0.5$$
