MATL, 20 19 17 bytes

2-:XdtP!+~TTYa1YG

You can try it experimentally in MATL online. You may need to refresh the page if it doesn't work.

Sample run:

ASCII version: 19 bytes

2-:XdtP!+~TTYa~42*c

Try it online!

JavaScript (ES6), 96 bytes

f=
n=>[...Array(n--)].map((_,i,a)=>a.map((_,j)=>i&&j&&n-i&&n-j&&i-j&&n-i-j?' ':'*').join``).join`
`
;

<input type=number min=1 step=2 oninput=  o.textContent=f(this.value)><pre id=o>

Java 7, 131 130 128 125 bytes

String c(int n){String r="";for(int i=-1,j;++i<n;r+="\n")for(j=0;j<n;r+=i<1|j<1|n-i<2|n-j<2|i==j|i==n-++j?"*":" ");return r;}

3 bytes saved thanks to @LeakyNun.

Ungolfed & test code:

Try it here.

class M{
  static String c(int n){
    String r = "";
    for(int i = -1, j; ++i < n; r += "\n"){
      for(j = 0; j < n;
            r += i < 1      // Responsible for the first horizontal line
               | j < 1      // Responsible for the first vertical line
               | n-i < 2    // Responsible for the last horizontal line
               | n-j < 2    // Responsible for the last vertical line
               | i == j     // Responsible for the top-left to bottom-right diagonal line
               | i == n-++j // Responsible for the top-right to bottom-left diagonal line (and increasing j)
             ? "*"
             : " ");
    }
    return r;
  }

  public static void main(String[] a){
    System.out.println(c(1));
    System.out.println(c(3));
    System.out.println(c(5));
    System.out.println(c(7));
  }
}

Output:

*

***
***
***

*****
** **
* * *
** **
*****

*******
**   **
* * * *
*  *  *
* * * *
**   **
*******

C#, 112 101 bytes

Thanks to TheLethalCoder for reminding me that these anonymous lambda statement-nor-expression things are allowed in C#.

n=>{var r="";for(int y=n--,x;y-->0;r+="*\n")for(x=0;x<n;r+=y%n*x<1|y==x|y==n-x++?"*":" ");return r;};

Who said C# isn't a fun golfing language?

Python, 114 110 96 bytes

Totally changed:

def b(n):m=n-1;g=range(n);return[bin(sum(2**p for p in 0<r<m and{0,m,r,m-r}or g))[2:]for r in g]

Returns a list of strings, characters using 1 and 0.

Test it at ideone

Previous Python 2 @110

def f(n):g=range(n);n-=1;print'\n'.join(''.join((c in(r,n-r,0,n)or r in(0,n))and'#'or' 'for c in g)for r in g)

Test it on ideone

Matlab, 68 66 64 58 bytes

Since graphical output is allowed too:

k=input('');[x,y]=ndgrid(abs(-k:k));spy(~(max(x,y)<k&x-y))

Which outputs e.g.

The ascii only versions would be:

This is using the indexing 0,1,2,3,...

k=input('');[x,y]=ndgrid(abs(-k:k));[(max(x,y)==k|~(x-y))*42,'']

Alternatively with the indexing 1,3,7,...:

n=input('');k=1:n;m=eye(n);m([k,end-k+1])=1;[(m|flip(m'))*42,'']

R, 102 bytes

    n=scan();for(i in 1:n){for(j in 1:n){z=" ";if(i%in%c(1,n,n-j+1)|j%in%c(1,i,n))z="*";cat(z)};cat("\n")}

Note that it is more efficient to express the condition using %in% than i==1|j==1|...

Java, 130 bytes

s->{for(int i=0;i<s;i++)for(int j=0;j<s;j++)System.out.print((s-1-i==j||i==j||i==0||j==0||i==s-1||j==s-1)?j==s-1?"*\n":"*":" ");};

Test Program

Consumer<Integer> consumer = s -> {
        for (int i = 0; i < s; i++) {
            for (int j = 0; j < s; j++) {
                System.out.print((s - 1 - i == j || i == j || i == 0 || j == 0 || i == s - 1 || j == s - 1) ? j == s - 1 ? "*\n" : "*" : " ");
            }
        }
    };

    consumer.accept(20);

Pyth, 27 25 bytes

jmsm?|qFKaLJ/Q2,dk}JKN\ Q

Try it online!

Probably golfable.

Haskell, 102 bytes 100 bytes

Saved 2 bytes, thanks to flawr.

c s=unlines$f(\m->f(m#))where f=(<$>[1..s]);m#n|or((==)<$>[n,m]<*>[1,s])||n==m||n==s-m+1='*'|1>0=' '

Ungolfed version:

cross :: Int -> String
cross s = unlines $ map line [1..s]
    where line m = map (pos m) [1..s]
          pos m n | n == m = '*'
                  | n == s - m + 1 = '*'
                  | elem m [1, s] = '*'
                  | elem n [1, s] = '*'
                  | otherwise = ' '

I'm sure this can still be improved, but this is what I've come up with for now.

Scala, 141 137 bytes

val s=args(0).toInt-1;val t=0 to s;print(t.map{x=>t.map{y=>if(x==0||x==s||y==0||y==s||x==y||x==s-y)"*" else " "}.mkString+"\n"}.mkString)

Run:

$ scala cross.scala 10

Technically I could remove the print stuff and go to something like

def c(n:Int)={val (s,t)=(n-1,0 to n-1);t.map{x=>t.map{y=>if(x==0||x==s||y==0||y==s||x==y||x==s-y)"*" else " "}.mkString+"\n"}.mkString}

This would make it 135 or 121 bytes depending on whether you count the function syntax stuff.

Readable version:

def cross(n: Int) = {
   // Declares both s and t as variables with tuple expansion
   // s is the zero-based size and t is a range from 0 to s
   val (s,t) = (n-1, 0 to n-1)

   // Maps all rows by mapping the columns to a star or a space
   t.map { x =>
      t.map { y =>
        if (x == 0 || x == s || y == 0 || y == s || x == y || x == s-y) "*" 
        else " "
      }.mkString+"\n" // Concatenate the stars and spaces and add a newline
   }.mkString         // Concatenate the created strings
 }

C, 140 121 114 bytes

19 bytes thanks to Quentin.

7 bytes saved by switching from a double-nested loop to one loop.

main(a){scanf("%d",&a);for(int i=0;i<a*a;i++,i%a||puts(""))putchar(i/a&&i/a^a-1&&i%a&&-~i%a&&i%-~a&&i%~-a?32:42);}

Golfing suggestions welcome.

Logo, 155 bytes

Graphical solution, implemented as a function

I retooled my answer for Alphabet Triangle and changed the angles around a bit. As before, r draws a line of characters. This time, the b function draws a box by drawing one straight edge and one diagonal, rotating, and repeating four times. This causes the diagonals to be drawn twice (on top of each other), but it was less code than handling it separately. This answer also properly handles even numbers. I had to add special handling for an input of 1 to prevent it from going forward.

I implemented it as a function, b which takes the size as an argument:

pu
to r:n:b:l repeat:n[rt:b label "A lt:b if repcount>1[fd:l]] end
to b:s
repeat 4[rt 90
r:s 90-heading 20 rt 135
r:s 90-heading 20*sqrt 2 rt 45]
end

Try it out on Calormen.com's Logo interpreter. To call it, append a line and call b in the following format:

b 7

... or try the sampler platter, which draws four samples in sizes 5, 7, 9, and 11, rotating 90 degrees in between:

repeat 4[
  b repcount*2+3
  rt 90
]

PowerShell (133)

filter s($x){1..$x|%{$o="";$r=$_;1..$x|%{if($_-eq1-or$r-eq1-or$_-eq$x-or$r-eq$x-or$r-eq$_-or$r-1-eq$x-$_){$o+="*"}else{$o+="_"}};$o}}

Clunky, but it works well enough.

s(11)
***********
**_______**
*_*_____*_*
*__*___*__*
*___*_*___*
*____*____*
*___*_*___*
*__*___*__*
*_*_____*_*
**_______**
***********

Golfing suggestions definitely welcome, it's been too long since I've PowerShell'd.

Python 2, 83 bytes

i=n=input()
while i:l=['* '[1<i<n]]*n;i-=1;l[0]=l[~0]=l[i]=l[~i]='*';print`l`[2::5]

Modifies a list of the row's characters to put a * into the first, last, i'th, and i'th-to-last place. The first and last row start as all *, and the rest as all spaces. Works for evens too. A lambda expression is probably shorter than modification, but I like this method.

Python 2, 65 bytes

i=n=2**input()/2
while i:print bin((n>i>1or~-n)|n|i|n/i)[2:];i/=2

Uses Jonathan Allan's idea of outputting binary numbers like:

11111
11011
10101
11011
11111

The rows are created with bit arithmetic and displayed in binary. Each part it or'ed into the rest. The part are are produced by powers of 2 n (fixed) and i (falling) via

1. Left side 1
2. Right side n
3. Diagonals i and n/i
4. Top and bottom by n-1 when i==1 or i==n.

Actually, (1) and (4) are combined by producing 1 when 1<i<n and n-1 otherwise.