Is the transpose of a g inverse the g inverse of transpose of the matrix?

#1
While working with linear models,I came across the situation where the proof of some results might be easier if we apply that transpose of g inverse is g inverse of transpose of a matrix.In R this codes
Code:
library(MASS)
x1=c(1,0,1,1)
x2=c(3,6,5,2)
x3=c(2,4,6,1)
x4=c(3,5,2,1)
m1=cbind(x1,x2,x3,x4)
ginv(m1)
m2=t(m1)
ginv(m2)
gave me
Code:
> ginv(m1)
       [,1]        [,2]        [,3]       [,4]
[1,]  0.625 -0.45833333  0.04166667  0.3333333
[2,] -0.875  0.37500000 -0.12500000  1.0000000
[3,]  0.375 -0.20833333  0.29166667 -0.6666667
[4,]  0.750 -0.08333333 -0.08333333 -0.6666667
> m2=t(m1)
> ginv(m2)
            [,1]   [,2]       [,3]        [,4]
[1,]  0.62500000 -0.875  0.3750000  0.75000000
[2,] -0.45833333  0.375 -0.2083333 -0.08333333
[3,]  0.04166667 -0.125  0.2916667 -0.08333333
[4,]  0.33333333  1.000 -0.6666667 -0.66666667
.But we know the g inverse of a matrix is not unique. Also we note the condition of g inverse is
\(AGA=A\)
\(A^tG^tA^t=A^t\)

So Shall we restrict ourselves by saying that if G is the G inverse of A then\( G^t\) is one of the g inverse of \(A^t \)? or more generally we can say that
\((A^-)^t=(A^t)^-\)
just like ordinary inverses?Also any link for studying this will be helpful.
 
#3
Thank you for your link.But that does not really contain my answer.Precisely my question is whether
\((A^-)^t=(A^t)^-\)
holds for any matrix A.