Let's break down the code step by step:

1. The function `my_function` takes an integer `n` as input.

2. The function checks if `n` is equal to 1. If it is, the function returns without doing anything.

3. If `n` is not 1, the function uses two nested for loops:

   - The outer loop iterates from 1 to `n` (inclusive).
   - The inner loop iterates from 1 to `n` (inclusive).

4. Inside the inner loop, it prints a single asterisk (`*`) followed by a newline (`\n`).

Now, let's analyze the time complexity using Big-O notation:

- The outer loop runs `n` times, and for each iteration, the inner loop runs `n` times.
- Therefore, the total number of iterations is `n * n`, which simplifies to `n^2`.

Since the number of operations (printing asterisks) is proportional to `n^2`, the time complexity is O(n^2).

Here's the corrected code with the additional newline character (`\n`) at the end of the inner loop:

```python
def my_function(n):
  if n == 1:
    return

  for i in range(1,n+1):
    for j in range(1, n+1):
      print("*", end="")
      print()  # Add a newline character

my_function(5)
```

When you run this code, it will print a 5x5 grid of asterisks, each on a new line.